Chapter 2

Exponents and Radicals

2.1 Exponents

Exponents are mathematical shorthand that tells us to multliply the same number by itself for a specific number of times. In this chapter, you will learn about positive integral exponent,zero exponent, negative integral exponents, rational exponents, radicals and exponential equations.

ထပ်ညွှန်းများသည်တိကျအရေအတွက်ရှိသည့်ကိန်းတစ်ခု၏မြှောက်လဒ်ကိုဖော်ပြသောသင်္ချာအတိုကောက်ဖြစ်သည်။ဤသင်ခန်းစာတွင်အပေါင်းကိန်းပြည့်ထပ်ညွှန်းများ၊သုညထပ်ညွှန်း၊အနုတ်ကိန်းပြည့်ထပ်ညွှန်းများ၊ရာရှင်နယ်(အပိုင်းကိန်း)ထပ်ညွှန်းများ၊ radicals (root ဖြင့်ဖော်ပြသောကိန်းများ)နှင့်ထပ်ညွှန်းညီမျှခြင်းများကိုလေ့လာကြမည်။


2.1.1 Positive Integral Exponents
In general, if $a$ is any real number and $n$ is a positive integer, then the $n^{th}$ power of $a$ is

$\underbrace{a \times a \times a \times a \times \cdots \times a}_{n \text { factors }}=a^{n}$

where the number $a$ is called the base and $n$ is called the exponent or index.

$a$ သည်ကိန်းစစ်တစ်ခုဖြစ်ပြီး $n$ သည်အပေါင်းကိန်းပြည့်တစ်ခုဖြစ်လျှင် $n^{th}$power of $a$ ဆိုသည်မှာ $a$ ကို $n$ အကြိမ်မြှောက်ထားခြင်းဖြစ်သည်ဟုဆိုလိုသည်။

$a$ ကို base (အခြေ)ဟုခေါ်သည်။ $n$ ကို exponent (သို့မဟုတ်)၊ index (ထပ်ညွှန်း)ဟုခေါ်သည်။

မြှောက်လဒ်တစ်ခုရေးသားဖော်ပြရာတွင်အမြှောက်သင်္ကေတ cross($\times$) အစား raised dot($\cdot$) ကိုလည်းအသုံးပြုလေ့ရှိသည်။

$3\times 3=3 \cdot 3=3^{2}$

$5\times 5\times 5=5 \cdot 5 \cdot 5=5^{3}$

$a \times a \times a \times a=a \cdot a \cdot a \cdot a=a^{4}$

2.1.2 Zero and Negative Integral Exponents
Definition: For any real number $x$, if $x\neq0$, then $x^{0}=1$.

For example,

$\begin{aligned} 3^{0}=1 \\\\ (-4)^{0}=1 \end{aligned}$

Note that $0^{0}$ is indeterminate.

Definition: For any real number $x$, if $x\neq0$, then $x^{n}=\dfrac{1}{x^{n}}$.

For example, $\begin{aligned} 5^{-2}&=\dfrac{1}{5^{2}}=\dfrac{1}{25} \\\\ \left(\dfrac{1}{3}\right)^{-4}&=\dfrac{1}{\left(\dfrac{1}{3}\right)^{4}}=\dfrac{1}{\dfrac{1}{3^{4}}}=3^{4} \end{aligned}$

Consequently, $\dfrac{1}{x^{-n}}=x^{n}$.

Example 1

Evluate the following.

(i)  $\dfrac{3^{-3}}{4^{-3}}$   (ii)  $5^{0}-5x^{0}-(5x)^{-1}-(5x)^{0}$


(i) $\dfrac{3^{-4}}{4^{-3}}=\dfrac{4^{3}}{3^{4}}=\dfrac{64}{81}$

(ii) $5^{0}-5 x^{0}-(5 x)^{-1}-(5 x)^{0}=1-5(1)-\dfrac{1}{5 x}-1=-5-\dfrac{1}{5 x}$


2.1.3 Rules for Integral Exponents
$x$ နှင့် $y$ တို့သည်ကိန်းစစ်များဖြစ်ပြီး $m$ နှင့် $n$ တို့သည်အပေါင်းကိန်းပြည့်များဖြစ်သည်ဟုယူဆလျှင်

Rule 1   (Multiplication)

$x^{m} \cdot x^{n}=x^{m+n}$

For example,

$\begin{aligned} 3^{3} \cdot 3^{5}&=3^{3+5}=3^{8} \\\\ x^{2} \cdot x^{6}&=3^{2+6}=x^{8} \end{aligned}$

Rule 2   (Division)

$x^{m} \div x^{n}=\dfrac{x^{m}}{x^{n}}$= $\begin{cases} x^{m-n}, & \text { if } m>n \\\\ 1, & \text { if } m=n \\\\ \dfrac{1}{x^{n-m},} & \text { if } m < n, x \neq 0 \end{cases}$

For example,

$\begin{aligned} 3^{5} \div 3^{2}&=3^{5-2}=3^{3} \\\\ a^{3} \div a^{8}&=\dfrac{a^{3}}{a^{8}}=\dfrac{1}{a^{8-3}}=\dfrac{1}{a^{5}} \\\\ a^{4} \div a^{4}&=a^{4-4}=a^{0}=1 \end{aligned}$

Rule 3   (Power of a Power)

$\left(x^{m}\right)^{n}=x^{m n}$

For example,

$\begin{aligned} \left(2^{3}\right)^{4}&=2^{3 \times 4}=2^{12} \\\\ \left(x^{4}\right)^{2}&=x^{4 \times 2}=x^{8} \end{aligned}$

Rule 4   (Power of a Product)

$(x y)^{n}=x^{n} \cdot y^{n}$

For example,

$\begin{aligned} (2 \cdot 3)^{4}&=2^{4} \cdot 3^{4} \\\\ (x y)^{5}&=x^{5} \cdot y^{5} \end{aligned}$

Rule 5   (Power of a Quotient)

$\left(\dfrac{x}{y}\right)^{n}=\dfrac{x^{n}}{y^{n}}, y \neq 0$

For example,

$\begin{aligned} \left(\dfrac{3}{5}\right)^{2}&=\dfrac{3^{2}}{5^{2}} \\\\ \left(\dfrac{x}{y}\right)^{3}&=\dfrac{x^{3}}{y^{3}}, y \neq 0 \end{aligned}$

Example 2

Simplify and name the rules used.

(i)  $x^{7}.x^{4}$   (ii)  $a^{15}÷a^{7}$   (iii)  $(b^{2})^{3}$

(iv)  $a^{5}.b^{5}$   (v)  $\dfrac{y^{7}}{4^{7}}$


(i)  $x^{7} \cdot x^{4}=x^{7+4}=x^{11}$  (Multiplication Rule)

(ii)  $a^{15} \div a^{7}=a^{15-7}=a^{8}$  (Division Rule)

(iii)  $\left(b^{2}\right)^{3}=b^{6} $  (Power of a Power Rule)

(iv)  $a^{5} \cdot b^{5}=(a b)^{5}$  (Power of a Product Rule)

(v)  $\dfrac{y^{7}}{y^{4}}=y^{7-4}=y^{3}$  (Power of a Quotient Rule)


Example 3

Simplify and name the rules used.

(i)  $\left(\dfrac{-81x^{3}y^{4}}{27xy^{3}}\right)^{3}$   (ii)  $(a^{-1}b^{-3})^{-2}$


$\begin{aligned} \text{ (i) }\quad \left(\dfrac{-81 x^{3} y^{4}}{27 x y^{3}}\right)^{3} &=\left(-3 x^{2} y\right)^{3}\quad \text { (Division Rule) } \\\\ &=(-3)^{2}\left(x^{2}\right)^{3} y^{3}\quad\text { (Power of a Power Rule) }\\\\ &=(-3)^{2}\left(x^{2}\right)^{3} y^{3}\quad \text {(Power of a Product Rule) }\\\\ &=-27 x^{6} y^{3} \quad \text { (Power of a Power Rule) } \end{aligned}$

$\begin{aligned} \text{ (ii) }\quad \left(a^{-1} b^{-3}\right)^{-2}&=\left(a^{-1}\right)^{-2}\left(b^{-3}\right)^{-2}\quad\text{ (Power of a Product Rule) }\\\\ &=a^{2} b^{6}\quad\text{ (Power of a Power Rule) } \end{aligned}$


Example 4

Evaluate $\left(\dfrac{16.27}{25}\right)^{2}\left(\dfrac{50}{36}\right)^{3}$.


$\begin{aligned} (\dfrac{16 \cdot 27}{25})^{2}(\dfrac{50}{36})^{3} &=(\dfrac{2^{4} \cdot 3^{3}}{5^{2}})^{2}(\dfrac{2 \cdot 5^{2}}{2^{2} \cdot 3^{2}})^{3} \\\\ &=\dfrac{(2^{4} \cdot 3^{3})^{2}}{(5^{2})^{2}} \dfrac{(2 \cdot 5^{2})^{3}}{(2^{2} \cdot 3^{2})^{3}} \\\\ &=\dfrac{2^{8} \cdot 3^{6}}{5^{4}} \dfrac{2^{3} \cdot 5^{6}}{2^{6} \cdot 3^{6}} \\\\ &=2^{5} \cdot 5^{2} \\\\ &=32 \cdot 25 \\\\ &=800 \end{aligned}$

Example 5

Evaluate $\dfrac{(x^{2}-y^{2})^{3}}{(x+y)^{3}}$.

$\begin{aligned} \dfrac{\left(x^{2}-y^{2}\right)^{3}}{(x+y)^{3}} &=\dfrac{((x+y)(x-y))^{3}}{(x+y)^{3}} \\\\ &=\dfrac{(x+y)^{3}(x-y)^{3}}{(x+y)^{3}} \\\\ &=(x-y)^{3} \end{aligned}$

2.1.4 Rational Exponents
ဤအပိုင်းတွင်ကိန်းတစ်ခု၏ကိန်းရင်းနှင့်ရာရှင်နယ်ထပ်ညွှန်းများအကြောင်းလေ့လာကြမည်။

Definition: If $n$ is a positive integer and $x$ and $y$ are real numbers, such that $x^{n}=y$, then $x$ is called the $n^{th}$ root of $y$.

$n$သည်အပေါင်းကိန်းပြည့်တစ်ခုဖြစ်၍ $x$ နှင့် $y$ တို့သည် $x^{n}=y$ ဖြစ်စေမည့်ကိန်းစစ်များဖြစ်ကြလျှင် $x$ ကို $y$ ၏ $n$ ထပ်ကိန်းရင်း $n^{th}$ root of $y$ ဟုခေါ်၏။

For example,

(i) $2^{3}=8$

2 is the cube root of 8.

$2$သည် $8$ ၏သုံးထပ်ကိန်းရင်းဖြစ်သည်။

(ii) $(-2)^{3}=-8$

$-2$ is the cube root of $-8$.

$-2$ သည် $-8$ ၏သုံးထပ်ကိန်းရင်းဖြစ်သည်။

(iii) $(-3)^{4}=81$

$-3$ is the fourth root of 81.

$-3$ သည် $81$ ၏လေးထပ်ကိန်းရင်းဖြစ်သည်။

(iv) $3^{4}=81$

3 is the fourth root of 81.

$3$ သည် $81$ ၏လေးထပ်ကိန်းရင်းဖြစ်သည်။

In general, for $x^{n}$, if $n$ is odd, there is only one real $n^{th}$ root of $y$, no matter whether $y$ is negative or zero or positive. In this case, the real $n^{th}$ root of $y$ is denoted by $\sqrt[n]{y}$ is called the principal $n^{th}$ root of $y$.

ယေဘုယျအားဖြင့် $n$ သည်အပေါင်း"မ"ကိန်းတစ်ခုဖြစ်ပြီး $x^{n}=y$ ဖြစ်လျှင် $y$ ၏ကိန်းစစ်ဖြစ်သည့် $n$ ထပ်ကိန်းရင်းသည်တစ်ခုသာရှိသည်။

$n^{th}$ root of $y$ ကို $\sqrt[n]{y}$ ဟုမှတ်သားပြီး the principal $n^{th}$ root of $y$ ဟုဖတ်သည်။

By (i) and (ii), $3$ is the cube root of $8$ and $-8$ and the principal third root of $8$ and $-8$, we write

$\begin{aligned} \sqrt[3]{8}&=2 \text{ (and) }\\\\ \sqrt[3]{-8}&=-2 \end{aligned}$

If $n$ is even and $y$ is positive, there are two real $n^{th}$ roots of $y$, one positive and other negative. In that case, the positive $n^{th}$ root of $y$ is denoted by $\sqrt[n]{y}$ is called the principal $n^{th}$ root of $y$.

ယေဘုယျအားဖြင့် $n$ သည်အပေါင်း"စုံ"ကိန်းတစ်ခုဖြစ်ပြီး $y>0$ နှင့် $x^{n}=y$ ဖြစ်လျှင် $y$ တွင်ကိန်းစစ်ဖြစ်သည့် $n$ ထပ်ကိန်းရင်းနှစ်ခုရှိ၏။တစ်ခုသည်အပေါင်းဖြစ်ပြီးကျန်တစ်ခုသည်အနုတ်ဖြစ်၏။

$y$ ၏အပေါင်းဖြစ်သည့် $n$ ထပ်ကိန်းရင်းကို $\sqrt[n]{y}$ ဖြင့်မှတ်သားပြီး $n^{th}$ root of $y$ ဟုဖတ်သည်။

By (iii) and (iv), $3$ and $-3$ are fourth roots of $81$ and the principal fourth root of $81$, $\sqrt[4]{81}=3$

2.1.4 Rational Exponents
Notice the following:

(i) $x^{2}=-4$ has no real number $x$ because the square of any nonzero real number is positive.

သုညမဟုတ်သောကိန်းစစ်တိုင်း၏နှစ်ထပ်ကိန်းများတန်ဖိုးသည်အပေါင်းကိန်းဖြစ်သောကြောင့် $x^{2}=-4$ ကိုပြေလည်စေသောကိန်းစစ် $x$ မရှိပါ။

(ii) It is important to understand the difference between $\sqrt[n]{-b}$ and $-\sqrt[n]{b}$.

$\sqrt[n]{-b}$ is the principal $n^{th}$ root of $(-b)$ and $-\sqrt[n]{b}$ is the negative principal $n^{th}$ root of $b$.

(iii) $\sqrt{x^{2}}= \begin{cases} x & \text { if } x \text { is positive or zero, } \\\\ -x & \text { if } x \text { is negative } \end{cases}$

For example,

(i) $\sqrt{9}=\sqrt{3^{2}}=3$

(ii) $\sqrt{9}=\sqrt{(-3)^{2}}=-(-3)=3$

Definition: For a real number $x$ and an integer $n(n ≥2)$, $x^{\frac{1}{n}}=\sqrt[n]{x}$, when $n$ is even, $x$ must be positive or zero.

ကိန်းစစ် $x$ နှင့် $n ≥2$ ဖြစ်သောကိန်းပြည့် $n$ အတွက် $x^{\frac{1}{n}}=\sqrt[n]{x}$ ဟုသတ်မှတ်သည်။

$n$ သည်စုံကိန်းတစ်ခုဖြစ်လျှင် $x$ သည်အပေါင်းကိန်း (သို့မဟုတ်)၊သုညဖြစ်ရမည်။

Definition: If $m$ and $n$ are positive integers and $\dfrac{m}{n}$ is a rational number in lowest terms, then for any real number $x$,

$x^{\frac{m}{n}}= \sqrt[n]{x^{m}}=(\sqrt[n]{x})^{m}$,

where $n$ is even, $x$ must be positive or zero.

$m$ နှင့် $n$ တို့သည်အပေါင်းကိန်းများဖြစ်ပြီး $\dfrac{m}{n}$ သည်အနိမ့်ဆုံးကျဉ်းပိုင်းရှိသောရာရှင်နယ်ကိန်းတစ်ခုဖြစ်လျှင် $x^{\frac{m}{n}}= \sqrt[n]{x^{m}}=(\sqrt[n]{x})^{m}$ ဟုသတ်မှတ်၏။ $n$ သည်စုံကိန်းတစ်ခုဖြစ်လျှင် $x$ သည်အပေါင်းကိန်း(သို့မဟုတ်)၊သုညဖြစ်ရမည်။

For example,

$(-8)^{\frac{2}{6}}=\sqrt[6]{(-8)^{2}}=\sqrt[6]{64}=2$

It is false because the exponent $\dfrac{2}{6}$ is not a lowest terms.

$(-8)^{\frac{2}{6}}=(-8)^{\frac{1}{3}}=\sqrt[3]{(-8)}=-2$  (True)

Definition: If $m$ and $n$ are any positive integers, then for any real number $x\neq0$, $x^{-\frac{m}{n}}=\dfrac{1}{x^{\frac{m}{n}}}$.

$m$ နှင့် $n$ တို့သည်အပေါင်းကိန်းများဖြစ်ပြီး $x$ သည်သုညမဟုတ်သောကိန်းစစ်တစ်ခုဖြစ်လျှင် $x^{-\frac{m}{n}}=\dfrac{1}{x^{\frac{m}{n}}}$ ဟုသတ်မှတ်သည်။

For example,

(i) $8^{-\frac{2}{3}}=\dfrac{1}{8^{\frac{2}{3}}}=\dfrac{1}{(\sqrt[3]{8})^{2}}=\dfrac{1}{2^{2}}=\dfrac{1}{4}$

(ii) $32^{-\frac{7}{5}}=\dfrac{1}{32^{\frac{7}{5}}}=\dfrac{1}{(\sqrt[5]{32})^{7}}=\dfrac{1}{2^{2}}=\dfrac{1}{128}$

Integral exponents အတွက်ဉပဒေသများသည် rational exponents အတွက်လည်းမှန်ပေသည်။

$x^{2}-y^{2}=(x+y)(x-y)$

$\begin{aligned} x-y&=\left(x^{\frac{1}{2}}\right)^{2}-\left(y^{\frac{1}{2}}\right)^{2}\\\\ &=\left(x^{\frac{1}{2}}+y^{\frac{1}{2}}\right)\left(x^{\frac{1}{2}}-y^{\frac{1}{2}}\right) \end{aligned}$

$\begin{aligned} x^{\frac{1}{2}}-y^{\frac{1}{2}}&=\left(x^{\frac{1}{4}}\right)^{2}-\left(y^{\frac{1}{4}}\right)^{2}\\\\ &=\left(x^{\frac{1}{4}}+y^{\frac{1}{4}}\right)\left(x^{\frac{1}{4}}-y^{\frac{1}{4}}\right) \end{aligned}$

$x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$

$\begin{aligned} x-y&=\left(x^{\frac{1}{3}}\right)^{3}-\left(y^{\frac{1}{3}}\right)^{3}\\\\ &=\left(x^{\frac{1}{3}}-y^{\frac{1}{3}}\right)\left(x^{\frac{2}{3}}+x^{\frac{1}{3}} y^{\frac{1}{3}}+y^{\frac{2}{3}}\right) \end{aligned}$

$x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)$

$\begin{aligned} x+y&=\left(x^{\frac{1}{3}}\right)^{3}+\left(y^{\frac{1}{3}}\right)^{3}\\\\ &=\left(x^{\frac{1}{3}}+y^{\frac{1}{3}}\right)\left(x^{\frac{2}{3}}-x^{\frac{1}{3}} y^{\frac{1}{3}}+y^{\frac{2}{3}}\right) \end{aligned}$

Example 6

Simplify

(a)  $16^{\frac{1}{2}}$   (b)  $8^{\frac{1}{2}}$   (c)  $(-27)^{\frac{1}{3}}$


(a) $16^{\frac{1}{2}}=\sqrt{16}=4$

(b) $8^{\frac{1}{3}}=\sqrt[3]{8}=2$

(c) $(-27)^{\frac{1}{3}}=\sqrt[3]{-27}=-3$


Example 7

Simplify and express the answer with positive exponents $\left(\dfrac{2^{m}\sqrt{2^{-m}}}{\sqrt{2^{m+1}}}\right)^{-2m}$.


$\begin{aligned} \left(\dfrac{2^{m} \sqrt{2^{-m}}}{\sqrt{2^{m+1}}}\right)^{-2 m} &=\left(\dfrac{2^{m} \cdot 2^{-\frac{m}{2}}}{2^{\frac{m+1}{2}}}\right)^{-2 m} \\\\ &=\left(2^{m-\frac{m}{2}-\frac{m+1}{2}}\right)^{-2 m} \\\\ &=\left(2^{-\frac{1}{2}}\right)^{-2 m} \\\\ &=2^{m} \end{aligned}$

Example 8

Simplify $\dfrac{x-y^{-1}}{(x^{\frac{1}{2}}+y^{-\frac{1}{4}})(x^{\frac{1}{4}}-y^{-\frac{1}{4}})}$.

$\begin{aligned} \dfrac{x-y^{-1}}{\left(x^{\frac{1}{2}}+y^{-\frac{1}{2}}\right)\left(x^{\frac{1}{4}}-y^{-\frac{1}{4}}\right)} &=\dfrac{\left(x^{\frac{1}{2}}\right)^{2}-\left(y^{-\frac{1}{2}}\right)^{2}}{\left(x^{\frac{1}{2}}+y^{-\frac{1}{2}}\right)\left(x^{\frac{1}{4}}-y^{-\frac{1}{4}}\right)} \\\\ &=\dfrac{\left(x^{\frac{1}{2}}+y^{-\frac{1}{2}}\right)\left(x^{\frac{1}{2}}-y^{\frac{1}{2}}\right)}{\left(x^{\frac{1}{2}}+y^{-\frac{1}{2}}\right)\left(x^{\frac{1}{4}}-y^{-\frac{1}{4}}\right)} \\\\ &=\dfrac{\left(x^{\frac{1}{2}}-y^{-\frac{1}{2}}\right)}{\left(x^{\frac{1}{4}}-y^{-\frac{1}{4}}\right)} \\\\ &=\dfrac{\left(x^{\frac{1}{4}}\right)^{2}-\left(y^{-\frac{1}{4}}\right)^{2}}{\left(x^{\frac{1}{4}}-y^{\frac{1}{4}}\right)} \\\\ &=\dfrac{\left(x^{\frac{1}{4}}+y^{-\frac{1}{4}}\right)\left(x^{\frac{1}{4}}-y^{-\frac{1}{4}}\right)}{\left(x^{\frac{1}{4}}-y^{-\frac{1}{4}}\right)} \\\\ &=x^{\frac{1}{4}}+y^{-\frac{1}{4}} \\\\ &=x^{\frac{1}{4}}+\frac{1}{\frac{1}{4}} \\\\ &=\dfrac{x^{\frac{1}{4}} y^{\frac{1}{4}}+1}{y^{\frac{1}{4}}} \end{aligned}$

2.2 Radicals
Definition: The symbol $\sqrt[n]{b}$ is called a radical, $\sqrt[n]{}$ is the radical sign, $n$ is the order or index, and $b$ is called the radicand.

$\sqrt[n]{b}$ ကို radical expression ဟုခေါ်သည်။

$\sqrt[n]{}$ ကို radical sign ဟုခေါ်သည်။ $b$ ကို radicand သို့မဟုတ် base ဟုခေါ်သည်။

$\sqrt{5}$,$\sqrt[3]{27}$,$\sqrt[5]{6}$ စသည်တို့သည် radical expression များဖြစ်ကြသည်။

2.2.1 Rules for Radicals

Rule 1

$(\sqrt[n]{x})^{n}=\sqrt[n]{x^{n}}=x$

For example,

$(\sqrt[3]{5})^{3}=5$

$\sqrt[4]{b^{4}}=b$

Reule 2

$\sqrt[n]{x} \sqrt[n]{y}=\sqrt[n]{x y}$

For example,

$\begin{aligned} \sqrt[3]{5} \cdot \sqrt[3]{20} &=\sqrt[3]{5 \cdot 20}=\sqrt[3]{100} \\\\ \sqrt[4]{x^{2}} \sqrt[4]{y} &=\sqrt[4]{x^{2} y} \end{aligned}$

Rule 3

$\sqrt[m]{\sqrt[n]{x}}=\sqrt[m n]{x}$

For example,

$\sqrt{\sqrt[3]{5}}=\sqrt[6]{5}$

$\sqrt[3]{\sqrt[4]{x^{5}}}=\sqrt[12]{x^{5}}$

Rule 4

$\sqrt[n]{\dfrac{x}{y}}=\dfrac{\sqrt[n]{x}}{\sqrt[n]{y}}, y \neq 0$

For example,

$\begin{aligned} \sqrt[3]{\dfrac{5}{9}} &=\dfrac{\sqrt[3]{5}}{\sqrt[3]{9}} \\\\ \sqrt[4]{\dfrac{x^{5}}{y^{3}}} &=\dfrac{\sqrt[4]{x^{5}}}{\sqrt[4]{y^{3}}} \end{aligned}$

Rule 5

(a) $\sqrt[n]{x^{m}}=\sqrt[k n]{x^{k m}}$

(b) $\sqrt[n]{x^{m}}=\sqrt[\frac{n}{k}] {x^{\frac{m}{k}}},k\neq0$

For example,

$\begin{aligned} \sqrt[3]{5}&=\sqrt[3 \times 2]{5^{2}}=\sqrt[6]{5^{2}} \\\\ \sqrt[5]{b^{3}}&=\sqrt[5 \times 3]{b^{3 \times 3}}=\sqrt[15]{b^{9}} \\\\ \sqrt[6]{25}&=\sqrt[6]{5^{2}}=\sqrt[3]{5} \\\\ \sqrt[12]{6^{3}}&=\sqrt[4]{6} \end{aligned}$

Example 9

Simplify the followings.

(a)  $\sqrt{27}$   (b)  $\sqrt[3] { 16 }$   (c)  $\sqrt {\sqrt[3] { 128 }}$

(d)  $\sqrt[3] {-32}$   (e)  $\sqrt{72}$


(a) $\sqrt{27}=\sqrt{9 \times 3}=\sqrt{3^{2} \times 3}=3 \sqrt{3}$.

(b) $\sqrt[3]{16}=\sqrt[3]{8 \times 2}=\sqrt[3]{2^{3} \times 2}=2 \sqrt[3]{2}$.

(c) $\sqrt{\sqrt[3]{128}}=\sqrt[6]{128}=\sqrt[6]{2^{6} \times 2}=2 \sqrt[6]{2}$.

(d) $\sqrt[5]{-32}=\sqrt[5]{(-2)^{5}}=-2$.

(e) $\sqrt{72}=\sqrt{36 \times 2}=6 \sqrt{2}$.


Example 10

Change the expression with the same radical and Simplify the radicands.

(a)  $5a\sqrt[3]{3}$   (b)  $\sqrt[3] { 5 }\sqrt{3}$

(c)  $\sqrt[4]{2} \sqrt[3] { 3 }$ (d)  $\sqrt{2x}\sqrt[3] {3y}$


(a) $5 a \sqrt[3]{3}=\sqrt[3]{(5 a)^{3}} \sqrt[3]{3}=\sqrt[3]{125 a^{3}} \sqrt[3]{3}=\sqrt[3]{375 a^{3}}$

(b) $\sqrt[3]{5} \sqrt{3}=\sqrt[6]{5^{2}} \sqrt[6]{3^{3}}=\sqrt[6]{25} \sqrt[6]{27}=\sqrt[6]{675}$

(c) $\sqrt[4]{2} \sqrt[3]{3}=\sqrt[2]{2^{3}} \sqrt[12]{3^{4}}=\sqrt[12]{8} \sqrt[12]{81}=\sqrt[12]{648}$

(d) $\sqrt{2 x} \sqrt[3]{3 y}=\sqrt[6]{(2 x)^{3}} \sqrt[6]{(3 y)^{2}}=\sqrt[6]{8 x^{3}} \sqrt[6]{9 y^{2}}=\sqrt[6]{72 x^{3} y^{2}}$


2.3 Operations with Radicals
Order တူသော radical နှစ်ခုကိုမြှောက်ရန် Rule 2 ကိုတိုက်ရိုက်အသုံးပြုနိုင်သည်။ Order မတူသော radical နှစ်ခုကိုမြှောက်ရာတွင် Rule 5 ကိုသုံး၍ထို radical တို့၏ orderများကို equavilant ဖြစ်အောင်ပြောင်းပြီးမှမြှောက်လဒ်ရှာနိုင်သည်။
Radicand လည်းတူ၊ index လည်းတူသော radical များကို similar radical ဟုခေါ်ပြီးဖြန့်ဝေရဂုဏ်သတ္တိသုံး၍ပေါင်းနိုင်၊နုတ်နိုင်သည်။
Dissimilar radical များကိုဖြေရှင်းရန်ပိုင်းခြေကို rationalize ပြုလုပ်ပြီးမှဖြေရှင်းနိုင်သည်။

ပိုင်းခြေကို၎င်း၏မိတ်ဖက်ကိန်းဖြင့်မြှောက်ခြင်းဖြင့်ရာရှင်နယ်ပြုလုပ်နိုင်သည်။

In rationlizing the denominator, we multliply it by its conjugate. This process is based on the fact that
$(a+b)(a-b)=a^{2}-b^{2}$.

အောက်ပါ factor တို့ကိုလေ့လာကြည့်ပါ။factor တစ်ခုသည်အခြား factor တစ်ခု၏ conjugate ဖြစ်သည်။

(1)  $a+\sqrt{b}$ and $a-\sqrt{b}$

(2)  $a+b\sqrt{c}$ and $a-b\sqrt{c}$

(3)  $\sqrt{a}+\sqrt{b}$ and $\sqrt{a}-\sqrt{b}$

(4)  $a\sqrt{b}+c\sqrt{d}$ and $a\sqrt{b}-c\sqrt{d}$

Example 11

Multliply $\sqrt{6}$ by $\sqrt{15}$ and express in the simplest form.

$\sqrt{6} \sqrt{15}=\sqrt{6\cdot15}=\sqrt{2\cdot3 \cdot 3\cdot5}=3 \sqrt{10}$

Example 12

Multliply $\sqrt[3]{12}$ by $\sqrt{10}$.


$\begin{aligned} \sqrt[3]{12} \cdot \sqrt{10} &=\sqrt[6]{12^{2}} \sqrt[6]{10^{3}} \\\\ &=\sqrt[6]{\left(2^{2} \cdot 3\right)^{2}(2 \cdot 5)^{3}} \\\\ &=\sqrt[6]{2^{4} \cdot 3^{2} \cdot 2^{3} \cdot 5^{3}} \\\\ &=\sqrt[6]{2^{6} \cdot 2 \cdot 3^{2} \cdot 5^{3}} \\\\ &=2 \sqrt[6]{2250} \end{aligned}$

Example 13

Simplify $\sqrt{3}(\sqrt{3}+\sqrt{8})$.

$\begin{aligned} \sqrt{3}(\sqrt{3}+\sqrt{8}) &=\sqrt{3} \sqrt{3}+\sqrt{3} \sqrt{8} \\\\ &=3+\sqrt{3 \cdot 2^{2} \cdot 2} \\\\ &=3+2 \sqrt{6} \end{aligned}$

Example 14

Simplify $\sqrt{200}+\sqrt{50}-\sqrt{18}$.

$\begin{aligned} \sqrt{200}+\sqrt{50}-\sqrt{18} &=\sqrt{2 \cdot 100}+\sqrt{2.25}-\sqrt{2.9} \\\\ &=\sqrt{2 \cdot 10^{2}}+\sqrt{2.5^{2}}-\sqrt{2.3^{2}}\\\\ &=10\sqrt{2}+5\sqrt{2}-3\sqrt{2}\\\\ &=(10+5-3)\sqrt{2}\\\\ &=12 \sqrt{2} \end{aligned}$

Example 15

Simplify $\sqrt{24}$+$\sqrt{\dfrac{2}{3}}$-$\sqrt[3]{\dfrac{2}{9}}$.

$\begin{aligned} \sqrt{24}+\sqrt{\dfrac{2}{3}}-\sqrt[3]{\dfrac{2}{9}} &=\sqrt{4 \cdot 6}+\dfrac{\sqrt{2}}{\sqrt{3}}-\dfrac{\sqrt[3]{2}}{\sqrt[3]{3^{2}}} \\\\ &=2 \sqrt{6}+\dfrac{\sqrt{2}}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}-\dfrac{\sqrt[3]{2}}{\sqrt[3]{3^{2}}} \times \dfrac{\sqrt[3]{3}}{\sqrt[3]{3}} \\\\ &=2 \sqrt{6}+\frac{\sqrt{6}}{3}-\frac{\sqrt[3]{6}}{3} \\\\ &=\left(2+\dfrac{1}{3}\right) \sqrt{6}-\dfrac{\sqrt[3]{6}}{3} \\\\ &=\dfrac{7}{3} \sqrt{6}-\dfrac{1}{3} \sqrt[3]{6} \end{aligned}$

Example 16

Simplify $\dfrac{\sqrt{2}-3\sqrt{3}}{2\sqrt{2}-\sqrt{3}}$.

$\begin{aligned} \dfrac{\sqrt{2}-3 \sqrt{3}}{2 \sqrt{2}-\sqrt{3}} &=\dfrac{\sqrt{2}-3 \sqrt{3}}{2 \sqrt{2}-\sqrt{3}} \times \dfrac{2 \sqrt{2}+\sqrt{3}}{2 \sqrt{2}+\sqrt{3}} \\\\ &=\dfrac{4+\sqrt{6}-6 \sqrt{6}-9}{(2 \sqrt{2})^{2}-\sqrt{3}^{2}} \\\\ &=\dfrac{-5 \sqrt{6}-5}{8-3} \\\\ &=\dfrac{5(-\sqrt{6}-1)}{5} \\\\ &=-\sqrt{6}-1 \end{aligned}$

2.4 Exponential Equations
The exponential equation is an equation in which a variable occurs in the exponent.
ထပ်ညွှန်းတွင်ကိန်းရှင်တစ်ခုပါဝင်သောညီမျှခြင်းကို exponential equation ဟုခေါ်သည်။
Some exponential equations: $3^{x}=81$ and $2^{5x-1}=32$

If $x^{n}=x^{m}$, then $n=m$ for $x\neq 0$ and $x\neq 1$.

Exponential Equation များကိုဖြေရှင်းရာတွင်
ညီမျှခြင်းနှစ်ဖက်စလုံးရှိကိန်းများကိုအခြေတူကိန်းတစ်ခုတည်းဖြစ်အောင်လုပ်ဆောင်ရမည်။
အခြေကိန်းသည် $0$ မဟုတ်လျှင်၊$1$ မဟုတ်လျှင်ထပ်ညွှန်းချင်းညီ၍မသိကိန်းတန်ဖိုးကိုရှာနိုင်သည်။

Example 17

Solve the equation $2^{3x-1} = 32$.

$\begin{aligned} 2^{3 x-1}&=32 \\\\ 2^{3 x-1}&=2^{5} \\\\ 3 x-1&=5 \\\\ 3 x&=6 \\\\ x&=2 \end{aligned}$

Example 18

Solve the equation $3^{2x+1}\cdot 27^{x-1} = 81$.

$\begin{aligned} 3^{2 x+1} \cdot 27^{x-1} &=81 \\\\ 3^{2 x+1} \cdot\left(3^{3}\right)^{x-1} &=3^{4} \\\\ 3^{2 x+1} \cdot 3^{3 x-3} &=3^{4} \\\\ 3^{2 x+1+3 x-3} &=3^{4} \\\\ 3^{5 x-2} &=3^{4} \\\\ 5 x-2 &=4 \\\\ 5 x &=6 \\\\ x &=\frac{6}{5} \end{aligned}$

Example 19

Solve the equation $2 \cdot 2^{2x}-7\cdot 2^{2x}-4= 0$.

$\begin{aligned} 2 \cdot 2^{2 x}-7 \cdot 2^{x}-4&=0\\\\ 2 \cdot\left(2^{x}\right)^{2}-7 \cdot 2^{x}-4&=0\\\\ \text{ Let } 2^{x}&=a\\\\ 2 a-7 a-4&=0\\\\ (2 a+1)(a-4)&=0\\\\ 2 a+1=0 \quad\text{ (or) }\quad a-4&=0\\\\ a=-\dfrac{1}{2}\quad\text{ (or) }\quad a&=4\\\\ 2^{x}=-\dfrac{1}{2} \quad\text{ (or) }\quad 2^{x}&=4\\\\ \text{ impossible } \quad\text{ (or) }\quad 2^{x}&=2^{2}\\\\ \therefore x&=2 \end{aligned}$

Problems

Exercise 2.1

  1. Simplify by using the rules of exponents and name the rules used.

  2. (a)  $\dfrac{36 a^{4} b^{5}}{100 a^{7} b^{3}}$   (d)  $\left(\dfrac{x^{4}}{y^{5}}\right)^{3}\left(\dfrac{y^{3}}{x^{2}}\right)^{2}$

    (b)  $\dfrac{27 a^{2} b^{5}}{\left(9 a^{2} b\right)^{2}}$  (e)  $\dfrac{2^{3^{2}}}{\left(2^{2}\right)^{3}}$

    (c)  $\left(\dfrac{-135 a^{4} b^{5} c^{6}}{315 a^{6} b^{7} c^{8}}\right)^{2}$


    (a)  $\dfrac{36 a^{4} b^{5}}{100 a^{7} b^{3}}=\dfrac{9 b^{2}}{25 a^{3}}$ (Division Rule)

    $\begin{aligned} \text{ (b) }\quad \dfrac{27 a^{2} b^{5}}{\left(9 a^{2} b\right)^{2}}&=\dfrac{27 a^{2} b^{5}}{9^{2}\left(a^{2}\right)^{2} b^{2}} \quad \text{ (Power of a Product Rule) }\\\\ &=\dfrac{27 a^{2} b^{5}}{81 a^{4} b^{2}} \quad \text { (Power of a Power Rule) } \\\\ &=\dfrac{b^{3}}{3 a^{2}} \quad \text { (Division Rule) }\\\\ \end{aligned}$

    $\begin{aligned} \text{ (c) }\quad \left(\dfrac{-135 a^{4} b^{5} c^{6}}{315 a^{6} b^{7} c^{8}}\right)^{2}&=\left(\dfrac{-3}{7 a^{2} b^{2} c^{2}}\right)^{2}\quad \text{ (Division Rule) }\\\\ &=\dfrac{(-3)^{2}}{\left(7 a^{2} b^{2} c^{2}\right)^{2}} \quad \text { (Power of a Quotient Rule) } \\\\ &=\dfrac{9}{7^{2}\left(a^{2}\right)^{2}\left(b^{2}\right)^{2}\left(c^{2}\right)^{2}} \quad\text { (Power of a Product Rule) } \\\\ &=\dfrac{9}{49 a^{4} b^{4} c^{4}} \quad \text { (Power of a Power Rule) }\\\\ \end{aligned}$

    $\begin{aligned} \text{ (d) }\quad \left(\dfrac{x^{4}}{y^{5}}\right)^{3}\left(\dfrac{y^{3}}{x^{2}}\right)^{2}&=\dfrac{\left(x^{4}\right)^{3}}{\left(y^{5}\right)^{3}} \cdot \dfrac{\left(y^{3}\right)^{2}}{\left(x^{2}\right)^{2}} \quad \text{ (Power of a Quotient Rule) }\\\\ &=\dfrac{x^{12}}{y^{15}} \cdot \dfrac{y^{6}}{x^{4}} \quad \text{ (Power of a Power Rule) }\\\\ &=\dfrac{x^{8}}{y^{9}}\quad \text{ (Division Rule) }\\\\ \end{aligned}$

    $\begin{aligned} \text{ (e) }\quad \dfrac{2^{3^{2}}}{\left(2^{2}\right)^{3}}&=\dfrac{2^{9}}{2^{6}} \quad \text{ (Power of a Product Rule) }\\\\ &=2^{3} \quad \text{ (Division Rale) }\\\\ &=8 \end{aligned}$


  3. Evaluate the followings.

  4. (a)  $\dfrac{54^{2} \times 12^{3} \times 64^{2}\left(3^{2} \times 4^{3} \times 5^{2}\right)^{3}}{\left(3^{2} \times 15 \times 20^{3}\right)^{4}}$

    (b)  $\left(\dfrac{343}{36}\right)^{3}\left(\dfrac{540}{56}\right)^{4}$

    (c)  $\left(\dfrac{33}{1056}\right)^{3}\left(\dfrac{768}{270}\right)^{4}\left(\dfrac{450}{48}\right)^{3}$


    $\begin{aligned} \text{ (a) }\quad \dfrac{54^{2} \times 12^{3} \times 64^{2}\left(3^{2} \times 4^{3} \times 5^{2}\right)^{3}}{\left(3^{2} \times 15 \times 20^{3}\right)^{4}} &=\dfrac{\left(2 \times 3^{3}\right)^{2} \times\left(2^{2} \times 3^{3} \times\left(2^{6}\right)^{2}\left(3^{2} \times\left(2^{2}\right)^{3} \times 5^{2}\right)^{3}\right.}{\left(3^{2} \times 3 \times 5 \times\left(2^{2} \times 5\right)^{3}\right)^{4}} \\\\ &=\dfrac{2^{2} \times 3^{6} \times 2^{6} \times 3^{3} \times 2^{12} \times 3^{6} \times 2^{18} \times 5^{6}}{\left(3^{3} \times 2^{6} \times 5^{4}\right)^{4}} \\\\ &=\dfrac{2^{38} \times 3^{15} \times 5^{6}}{2^{24} \times 3^{12} \times 5^{16}} \\\\ &=\dfrac{2^{14} \times 3^{3}}{5^{10}} \end{aligned}$

    $\begin{aligned} \text { (b) }\quad\left(\dfrac{343}{36}\right)^{3}\left(\dfrac{540}{56}\right)^{4} &=\left(\dfrac{343}{36}\right)^{3}\left(\dfrac{135}{14}\right)^{4} \\\\ &=\left(\dfrac{7^{3}}{3^{2} \cdot 2^{2}}\right)^{3}\left(\dfrac{3^{3} \cdot 5}{7 \cdot 2}\right)^{4} \\\\ &=\dfrac{7^{9}}{3^{6} \cdot 2^{6}} \dfrac{3^{12} \cdot 5^{4}}{7^{4} \cdot 2^{4}} \\\\ &=\dfrac{7^{5} \cdot 3^{6} \cdot 5^{4}}{2^{10}} \end{aligned}$

    $ \begin{aligned} \text { (c) }\quad\left(\dfrac{33}{1056}\right)^{3}\left(\dfrac{768}{270}\right)^{4}\left(\dfrac{450}{48}\right)^{3} &=\left(\dfrac{1}{32}\right)^{3}\left(\dfrac{128}{45}\right)^{4}\left(\dfrac{75}{8}\right)^{3} \\\\ &=\left(\dfrac{1}{2^{5}}\right)^{3}\left(\dfrac{2^{7}}{3^{2} \times 5}\right)^{4}\left(\dfrac{3 \times 5^{2}}{2^{3}}\right)^{3} \\\\ &=\dfrac{1}{2^{15}} \dfrac{2^{28}}{3^{8} \cdot 5^{4}} \dfrac{3^{3} \cdot 5^{6}}{2^{9}} \\\\ &=\dfrac{2^{4} \cdot 5^{2}}{3^{5}} \\\\ &=\dfrac{16 \times 25}{243} \\\\ &=\dfrac{400}{243} \\\\ &=1 \dfrac{157}{243} \end{aligned}$

  5. Simplify.

  6. (a)  $\left(\dfrac{3^{m}}{15^{n}}\right)^{3}\left(\dfrac{45^{n}}{255^{m}}\right)^{2}$

    (b)  $\left(\dfrac{20^{x}}{400^{y}}\right)^{2}\left(\dfrac{150^{y^{2}}}{180^{x}}\right)^{3}$

    (c)  $\dfrac{\left(x^{3}-y^{3}\right)(x+y)}{\left(x^{2}-y^{2}\right)^{3}}$

    (d)  $\dfrac{\left(x^{a-b} x^{b-c}\right)^{a}\left(\dfrac{x^{a}}{x^{c}}\right)^{c}}{\left(x^{b} \cdot x^{c}\right)^{a} \div\left(x^{a+c}\right)^{c}}$


    $\begin{aligned} \text { (a) }\quad\left(\dfrac{3^{m}}{15^{n}}\right)^{3}\left(\dfrac{45^{n}}{255^{m}}\right)^{2}&=\left(\dfrac{3^{m}}{(3 \cdot 5)^{n}}\right)^{3}\left(\dfrac{\left(3^{2} \cdot 5\right)^{n}}{(3.5 \cdot 17)^{m}}\right)^{2}\\\\ &=\left(\dfrac{3^{m}}{3^{n} \cdot 5^{n}}\right)^{3}\left(\dfrac{3^{2 n} \cdot 5^{n}}{3^{m} \cdot 5^{m} \cdot 17^{m}}\right)^{2}\\\\ &=\dfrac{3^{3 m}}{3^{3 n} \cdot 5^{3 n}} \cdot \dfrac{3^{4 n} \cdot 5^{2 n}}{3^{2 m} \cdot 5^{2 m} \cdot 17^{2 m}}\\\\ &=\dfrac{3^{m} \cdot 3^{n}}{5^{n+2 m} \cdot 17^{2 m}}\\\\ &=\dfrac{3^{m+n}}{5^{n+2 m} 17^{2 m}} \end{aligned}$

    $\begin{aligned} \text { (b) }\quad\left(\dfrac{20^{x}}{400^{y}}\right)^{2}\left(\dfrac{150^{y^{2}}}{180^{x}}\right)^{3}&=\dfrac{20^{2 x}}{400^{2 y}} \cdot \dfrac{150^{3 y^{2}}}{180^{3 x}}\\\\ &=\dfrac{\left(2^{2} \cdot 5\right)^{2 x}}{\left(2^{4} \cdot 5^{2}\right)^{2 y}} \cdot \dfrac{\left(2 \cdot 3 \cdot 5^{2}\right)^{3 y^{2}}}{\left(2^{2} \cdot 3^{2} \cdot 5\right)^{3 x}}\\\\ &=\dfrac{2^{4 x} \cdot 5^{2 x}}{2^{8 y} \cdot 5^{4 y}} \cdot \dfrac{2^{3 y} \cdot 3^{3 y} \cdot 5^{6 y^{2}}}{2^{6 x} \cdot 3^{6 x} \cdot 5^{3 x}}\\\\ &=2^{4 x-6 x+3 y^{2}-8 y} \cdot 3^{3 y^{2}-6 x} \cdot 5^{2 x-3 x+6 y^{2}-4 y}\\\\ &=2^{3 y^{2}-8 y-2 x} \cdot 3^{3 y^{2}-6 x} \cdot 5^{6 y^{2}-4 y-x} \end{aligned}$

    $\begin{aligned} \text { (c) }\quad \dfrac{\left(x^{3}-y^{3}\right)(x+y)}{\left(x^{2}-y^{2}\right)^{3}} &=\dfrac{(x-y)\left(x^{2}+x y+y^{2}\right)(x+y)}{((x+y)(x-y))^{3}} \\\\ &=\dfrac{(x-y)\left(x^{2}+x y+y^{2}\right)(x+y)}{(x+y)^{3}(x-y)^{3}} \\\\ &=\dfrac{x^{2}+x y+y^{2}}{(x+y)^{2}(x-y)^{2}} \end{aligned}$

    $\begin{aligned} \text { (d) }\quad\dfrac{\left(x^{a-b} x^{b-c}\right)^{a}\left(\dfrac{x^{a}}{x^{c}}\right)^{c}}{\left(x^{b} \cdot x^{c}\right)^{a} \div\left(x^{a+c}\right)^{c}}&= \dfrac{\left(x^{a-b+b-c}\right)^{a}\left(x^{a-c}\right)^{c}}{\left(x^{b+c}\right)^{a} \div x^{a c+c^{2}}} \\\\ &=\dfrac{\left(x^{a-c}\right)^{a} \cdot x^{a c-c^{2}}}{a b+a c-a c-c^{2}} \\\\ &=\dfrac{x^{a^{2}-a c+a c-c^{2}}} \\\\ &=\dfrac{x^{a b-c}}{a^{2}-c^{2}} \\\\ &=x^{a^{2}-c^{2}-a b+c^{2}} \\\\ &=x^{a^{2}-a b} \end{aligned}$


  7. Evaluate the followings.

  8. (a)  $(-3)^{-2}$  (b)  $-3^{-3}$  (c)  $-2^{0}+5^{-1}$

    (d) $(-2)^{-3}+2^{-2}-2^{-4}$

    (e) $5^{0}-(-3)^{0}$

    (f) $\dfrac{27^{-6}}{125^{-3}} \div \dfrac{9^{-2}}{25^{-4}}$

    (g) $(-5)^{0}-(-5)^{-1}-(-5)^{-2}-(-5)^{-3}$

    (h) $(-1)^{(-1)^{-1}}$

    (i) $\dfrac{\left(180^{2}\right)^{-3}\left(6 \cdot 90^{-2}\right)^{3}}{\left(40^{-3}\right)^{2} \cdot 25^{-2}}$

    (j) $\dfrac{\left(2^{-3}-3^{-2}\right)^{-1}}{\left(2^{-3}+3^{-2}\right)^{-1}}$


    (a)  $(-3)^{-2}=\dfrac{1}{(-3)^{2}}=\dfrac{1}{9}$

    (b)  $-3^{-3}=-\dfrac{1}{3^{3}}=-\dfrac{1}{27}$

    (c)  $-2^{0}+5^{-1}=-1+\dfrac{1}{5}$

    $\begin{aligned} \text { (d) }\quad (-2)^{-3}+2^{-2}-2^{-4}&=\dfrac{1}{(-2)^{3}}+\dfrac{1}{2^{2}}-\dfrac{1}{2^{4}}\\\\ &=\dfrac{1}{-8}+\dfrac{1}{4}-\dfrac{1}{16}\\\\ &=\dfrac{-2+4-1}{16} \\\\ &=\dfrac{1}{16} \end{aligned}$

    (e)  $5^{0}-(-3)^{0}=1-1=0$

    $\begin{aligned} \text { (f) }\quad\dfrac{27^{-6}}{125^{-3}} \div \dfrac{9^{-2}}{25^{-4}} &=\dfrac{\left(3^{3}\right)^{-6}}{\left(5^{3}\right)^{-3}} \times \dfrac{25^{-4}}{9^{-2}} \\\\ &=\dfrac{3^{-18}}{5^{-9}} \times \dfrac{\left(5^{2}\right)^{-4}}{\left(3^{2}\right)^{-2}} \\\\ &=\dfrac{3^{-18}}{5^{-9}} \times \dfrac{5^{-8}}{3^{-4}}=\dfrac{5}{3^{14}} \end{aligned}$

    $\begin{aligned} \text { (g) }\quad(-5)^{0}-(-5)^{-1}-(-5)^{-2}-(-5)^{-3} &=1-\dfrac{1}{(-5)}-\dfrac{1}{(-5)^{2}}-\dfrac{1}{(-5)^{3}} \\\\ &=1+\dfrac{1}{5}-\dfrac{1}{25}+\dfrac{1}{125} \\\\ &=\dfrac{125+25-5+1}{125} \\\\ &=\dfrac{164}{125} \\\\ &=1 \dfrac{21}{125} \end{aligned}$

    (h) $(-1)^{(-1)^{-1}=(-1)^{\frac{1}{-1}}=(-1)^{-1}} =\dfrac{1}{-1}=-1$

    $\begin{aligned} \text { (i) }\quad\dfrac{\left(180^{2}\right)^{-3}\left(6 \cdot 90^{-2}\right)^{3}}{\left(40^{-3}\right)^{2} \cdot 25^{-2}} &=\dfrac{180^{-6} \cdot 6^{3} \cdot 90^{-6}}{40^{-6} \cdot 25^{-2}} \\\\ &=\dfrac{\left(2^{2} \cdot 3^{2} \cdot 5\right)^{-6} \cdot(2 \cdot 3)^{3} \cdot\left(2 \cdot 3^{2} \cdot 5\right)^{-6}}{\left(2^{3} \cdot 5\right)^{-6} \cdot\left(5^{2}\right)^{-2}} \\\\ &=\dfrac{2^{-12} \cdot 3^{-12} \cdot 5^{-6} \cdot 2^{3} \cdot 3^{2} \cdot 2^{-6} \cdot 3^{-12} \cdot 5^{-6}}{2^{-18} \cdot 5^{-6} \cdot 5^{-4}} \\\\ &=\dfrac{2^{-15} \cdot 3^{-21} \cdot 5^{-12}}{2^{-18} \cdot 5^{-10}} \\\\ &=\dfrac{2^{3}}{3^{21} \cdot 5^{2}} \end{aligned}$

    $ \begin{aligned} \text { (j) }\quad\dfrac{\left(2^{-3}-3^{-2}\right)^{-1}}{\left(2^{-3}+3^{-2}\right)^{-1}} &=\dfrac{2^{-3}+3^{-2}}{2^{-3}-3^{-2}} \\\\ &=\dfrac{\dfrac{1}{2^{3}}+\dfrac{1}{3^{2}}}{\dfrac{1}{2^{3}}-\dfrac{1}{3^{2}}} \\\\ &=\dfrac{\dfrac{1}{8}+\dfrac{1}{9}}{\dfrac{1}{8}-\dfrac{1}{9}} \\\\ &=\dfrac{\dfrac{9+8}{72}}{\dfrac{9-8}{72}} \\\\ &=\dfrac{17}{72} \times \dfrac{72}{1} \\\\ &=17 \end{aligned}$


  9. Simplify the followings.

  10. (a)  $\left(-3 a^{4}\right)\left(4 a^{-7}\right)$  (b)  $\left(\dfrac{2 x^{-4}}{5 y^{2} z^{3}}\right)^{-2}$

    (c)  $\left(\dfrac{x^{2 m+n} x^{3(m-n)}}{x^{m-2 n} x^{2 m-n}}\right)^{-3}$

    (d) $\left(\dfrac{2 x^{-3} y^{2}}{3^{-1} y^{3}}\right)^{2}\left(\dfrac{4 x^{-2} y^{3}}{3 x^{5}}\right)^{3} \div\left(\dfrac{81 x^{-2}}{y^{-3}}\right)^{-2}$

    (e) $\dfrac{2 x+y}{x^{-1}+2 y^{-1}}$  (f) $\left(x^{-2}-y^{-1}\right)^{-3}$

    (g)$\dfrac{x^{-2}-y^{-2}}{x^{-1}+y^{-1}}$   (h) $\dfrac{\left(x+y^{-1}\right)^{2}}{1+x^{-1} y^{-1}}$


    (a)  $\left(-3 a^{4}\right)\left(4 a^{-7}\right)=-12 a^{-3}=-\dfrac{12}{a^{3}}$

    $\begin{aligned} \text { (b) }\quad \left(\dfrac{2 x^{-4}}{5 y^{2} z^{3}}\right)^{-2}&=\dfrac{2^{-2} x^{8}}{5^{-2} y^{-4} z^{-6}}\\\\ &=\dfrac{5^{2} x^{8} y^{4} z^{6}}{2^{2}}\\\\ &=\dfrac{25 x^{8} y^{4} z^{6}}{4} \end{aligned}$

    $\begin{aligned} \text { (c) }\quad \left(\dfrac{x^{2 m+n} x^{3(m-n)}}{x^{m-2 n} x^{2 m-n}}\right)^{-3} &=\left(\dfrac{x^{2 m+n} x^{3 m-3 n}}{x^{m-2 n+2 m-n}}\right)^{-3} \\\\ &=\left(\dfrac{x^{3 m-2 n}}{x^{3 m-3 n}}\right)^{-3} \\\\ &=\dfrac{x^{-15 m+6 n}}{x^{-9 m+9 n}} \\\\ &=x^{-6 m-3 n} \end{aligned}$

    $\begin{aligned} \text { (d) }\quad &\left(\dfrac{2 x^{-3} y^{2}}{3^{-1} y^{3}}\right)^{2}\left(\dfrac{4 x^{-2} y^{3}}{3 x^{5}}\right)^{3} \div\left(\dfrac{81 x^{-2}}{y^{-3}}\right)^{-2} \\\\ &=\left(\dfrac{2 \cdot 3}{x^{3} y}\right)^{2}\left(\dfrac{2^{2} y^{3}}{3 x^{7}}\right)^{3} \div \dfrac{\left(3^{4}\right)^{-2} x^{4}}{y^{6}} \\\\ &=\dfrac{2^{2} \cdot 3^{2}}{x^{6} \cdot y^{2}} \times \dfrac{2^{6} \cdot y^{9}}{3^{3} \cdot x^{21}} \times \dfrac{y^{6}}{3^{-8} \cdot x^{4}} \\\\ &=\dfrac{2^{8} \cdot 3^{7} \cdot y^{13}}{x^{31}} \end{aligned}$

    $\begin{aligned} \text { (e) }\quad \dfrac{2 x+y}{x^{-1}+2 y^{-1}}&=\dfrac{2 x+y}{\dfrac{1}{x}+\dfrac{2}{y}}\\\\ &=\dfrac{2 x+y}{\dfrac{y+2 x}{x y}} \\\\ &=2 x+y \times \dfrac{x y}{(y+2 x)} \end{aligned}$

    $\begin{aligned} \text { (f) }\quad \left(x^{-2}-y^{-1}\right)^{-3}&=\left(\dfrac{1}{x^{2}}-\dfrac{1}{y}\right)^{-3}\\\\ &=\left(\dfrac{y-x^{2}}{x^{2} y}\right)^{-3}\\\\ &=\left(\dfrac{x^{2} y}{y-x^{2}}\right)^{3}\\\\ &=\dfrac{x^{6} y^{3}}{\left(y-x^{2}\right)^{3}} \end{aligned}$

    $\begin{aligned} \text { (g) } \quad \dfrac{x^{-2}-y^{-2}}{x^{-1}+y^{-1}} &=\dfrac{\frac{1}{x^{2}}-\dfrac{1}{y^{2}}}{\dfrac{1}{x}+\frac{1}{y}} \\\\ &=\dfrac{\dfrac{y^{2}-x^{2}}{x^{2} y^{2}}}{\dfrac{y+x}{x y}} \\\\ &=\dfrac{\dfrac{(y+x)(y-x)}{x^{2} y^{2}}}{\dfrac{y+x}{x y}} \\\\ &=\dfrac{(y+x)(y-x)}{x^{2} y^{2}} \times \dfrac{x y}{y+x} \\\\ &=\dfrac{y-x}{x y} \end{aligned}$

    $\begin{aligned} \text { (h) } \dfrac{\left(x+y^{-1}\right)^{2}}{1+x^{-1} y^{-1}} &=\dfrac{\left(x+\dfrac{1}{y}\right)^{2}}{1+\dfrac{1}{x y}} \\\\ &=\dfrac{\left(\dfrac{x y+1}{y}\right)^{2}}{\dfrac{x y+1}{x y}} \\\\ &=\dfrac{(x y+1)^{2}}{y^{2}} \times \dfrac{x y}{x y+1} \\\\ &=\dfrac{x y+1}{y} \end{aligned}$


Exercise 2.2

  1. Evaluate the followings.

  2. (a)  $(125)^{\frac{2}{3}}$  (b)  $(81)^{-\frac{3}{2}}$  (c)  $(-27)^{\frac{2}{3}}$

    (d) $\left(\dfrac{16}{81}\right)^{-\frac{3}{4}}$   (e) $\left(\dfrac{-125}{8} \div \dfrac{1}{64}\right)^{\frac{1}{3}}$

    (f) $(0.125)^{-\frac{2}{3}}=\left((0.5)^{3}\right)^{-\frac{2}{3}}$

    (g) $\left(\dfrac{64}{27}\right)^{-\frac{2}{3}}$

    (h) $(-4)^{-1}+(-1)^{-4}$


    (a) $(125)^{\frac{2}{3}}=\left(5^{3}\right)^{\frac{2}{3}}=5^{2}=25$

    (b) $(81)^{-\frac{3}{2}}=\left(9^{2}\right)^{-\frac{3}{2}}=9^{-3}=\dfrac{1}{9^{3}}=\dfrac{1}{729}$

    (c) $(-27)^{\frac{2}{3}}=\left((-3)^{3}\right)^{\frac{2}{3}}=(-3)^{2}=9$

    (d) $\left(\dfrac{16}{81}\right)^{-\frac{3}{4}}=\left(\left(\dfrac{2}{3}\right)^{4}\right)^{-\frac{3}{4}}=\left(\dfrac{2}{3}\right)^{-3}=\left(\dfrac{3}{2}\right)^{3}=\dfrac{27}{8}$

    (e) $\left(\dfrac{-125}{8} \div \dfrac{1}{64}\right)^{\frac{1}{3}}=\left(\dfrac{-125}{8} \times 6-4\right)^{\frac{1}{3}}=\left((-5)^{3} 2^{3}\right)^{\frac{1}{3}}=(-5) \times 2=-10$

    (f) $(0.125)^{-\frac{2}{3}}=\left((0.5)^{3}\right)^{-\frac{2}{3}}=(0.5)^{-2}=\dfrac{1}{(0.5)^{2}}=\dfrac{1}{0.25} \times \dfrac{100}{100}=\dfrac{100}{25}=4$

    (g) $\left(\dfrac{64}{27}\right)^{-\frac{2}{3}}=\left(\dfrac{4^{3}}{3^{3}}\right)^{-\frac{2}{3}}=\left(\dfrac{4}{3}\right)^{-2}=\left(\dfrac{3}{4}\right)^{2}=\dfrac{9}{16}$

    (h) $(-4)^{-1}+(-1)^{-4}=\dfrac{1}{-4}+\dfrac{1}{(-1)^{4}}=\dfrac{-5}{4}$


  3. Simplify the following.

  4. (a)  $\sqrt[3]{4^{2}} \cdot 4^{\frac{2}{3}} \cdot\left(\dfrac{1}{4}\right)^{-\frac{2}{3}}$

    (b)  $\sqrt{\dfrac{512 \times 27^{-3} \times 81 \times 3^{8}}{3^{4}}}$

    (c)  $\left(\left(\dfrac{3}{4}\right)^{-4}\right)^{-0.5} \cdot \sqrt{\left(\dfrac{4}{3}\right)^{-1}} \div 16^{-0.5}$

    (d)  $(27)^{\frac{1}{4}}+\dfrac{24}{(8)^{-\frac{2}{3}}}+\dfrac{\sqrt[5]{2}}{(4)^{\frac{-2}{5}}}$

    (e)  $ \dfrac{(243)^{\frac{4}{5}}+(64)^{\frac{2}{3}}-(216)^{\frac{1}{3}}}{(225)^{\frac{1}{2}}-(16)^{\frac{3}{4}}}$


    $\begin{aligned} \text { (a) } \quad\sqrt[3]{4^{2}} \cdot 4^{\frac{2}{3}} \cdot\left(\dfrac{1}{4}\right)^{-\frac{2}{3}}&=4^{\frac{2}{3}} \cdot 4^{\frac{2}{3}} \cdot 4^{\frac{2}{3}}\\\\ &=4^{\frac{2}{3}+\frac{2}{3}+\frac{2}{3}}\\\\ &=4^{\frac{6}{3}}\\\\ &=4^{2}\\\\ &=16 \end{aligned}$

    $\begin{aligned} \text { (b) } \quad\sqrt{\dfrac{512 \times 27^{-3} \times 81 \times 3^{8}}{3^{4}}} &=\sqrt{\dfrac{2^{9} \times 34 \times 3^{8}}{27^{3} \times 3^{4}}} \\\\ &=\sqrt{\dfrac{2^{8} \times 2 \times 3^{12}}{3^{9} \times 3^{4}}} \\\\ &=2 \times \sqrt{\dfrac{2}{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\\\ &=16 \dfrac{\sqrt{6}}{3} \end{aligned}$

    $\begin{aligned} \text { (c) } \quad\left(\left(\dfrac{3}{4}\right)^{-4}\right)^{-0.5} \cdot \sqrt{\left(\dfrac{4}{3}\right)^{-1}} \div 16^{-0.5} &=\left(\dfrac{3}{4}\right)^{2} \cdot \sqrt{\dfrac{3}{4}} \div\left(4^{2}\right)^{-0.5} \\\\ &=\dfrac{3^{2}}{4^{2}} \cdot \dfrac{\sqrt{3}}{2} \div 4^{-1} \\\\ &=\dfrac{9 \sqrt{3}}{32} \div \dfrac{1}{4} \\\\ &=\dfrac{9 \sqrt{3}}{32} \times 4 \\\\ &=\dfrac{9 \sqrt{3}}{8} \end{aligned}$

    $\begin{aligned} \text { (d) } \quad(27)^{\frac{1}{4}}+\dfrac{24}{(8)^{-\frac{2}{3}}}+\dfrac{\sqrt[5]{2}}{(4)^{\frac{-2}{5}}} &=3^{\frac{3}{4}}+24 \cdot 8^{\frac{2}{3}}+2^{\frac{1}{5}} \cdot 4^{\frac{2}{5}} \\\\ &=3^{\frac{3}{4}}+24 \cdot\left(2^{3}\right)^{\frac{2}{3}}+2^{\frac{1}{5}} \cdot\left(2^{2}\right)^{\frac{2}{5}} \\\\ &=3^{\frac{3}{4}}+96+2^{\frac{1}{5}} \cdot 2^{\frac{4}{5}} \\\\ &=98+3^{\frac{3}{4}} \end{aligned}$

    $\begin{aligned} \text { (e) } \quad \dfrac{(243)^{\frac{4}{5}}+(64)^{\frac{2}{3}}-(216)^{\frac{1}{3}}}{(225)^{\frac{1}{2}}-(16)^{\frac{3}{4}}} &=\dfrac{\left(3^{5}\right)^{\frac{4}{5}}+\left(4^{3}\right)^{\frac{2}{3}}-\left(6^{3}\right)^{\frac{1}{3}}}{\left(15^{2}\right)^{\frac{1}{2}}-\left(2^{4}\right)^{\frac{3}{4}}} \\\\ &=\dfrac{3^{4}+4^{2}-6}{15-2^{3}} \\\\ &=\dfrac{81+16-6}{7} \\\\ &=\dfrac{91}{7} \\\\ &=13 \end{aligned}$


  5. Simplify the following.

  6. (a)  $\dfrac{x-5 \sqrt{x}}{x-2 \sqrt{x}-15} \div\left(1+\dfrac{3}{\sqrt{x}}\right)^{-1}$

    (b)  $\sqrt[a]{\dfrac{\sqrt[b]{x}}{\sqrt[c]{x}}} \cdot \sqrt[b]{\dfrac{\sqrt[c]{x}}{\sqrt[a]{x}}} \cdot \sqrt[c]{\dfrac{\sqrt[a]{x}}{\sqrt[b]{x}}} $

    (c)  $\left(\dfrac{x^{m}-y^{m}}{x^{\frac{m}{2}}-y^{\frac{m}{2}}}-\frac{x^{m}-y^{m}}{x^{\frac{m}{2}}+y^{\frac{m}{2}}}\right)^{-2} $


    $\begin{aligned} \text { (a) }\quad \dfrac{x-5 \sqrt{x}}{x-2 \sqrt{x}-15} \div\left(1+\dfrac{3}{\sqrt{x}}\right)^{-1} &= \dfrac{\sqrt{x}^{2}-5 \sqrt{x}}{\sqrt{x}^{2}-2 \sqrt{x}-15} \div\left(\dfrac{\sqrt{x}+3}{\sqrt{x}}\right)^{-1} \\\\ &= \dfrac{\sqrt{x}(\sqrt{x}-5)}\\\\ &=(\sqrt{x}-5)(\sqrt{x}+3) \\\\ &= \dfrac{\sqrt{x}+3}{\sqrt{x}} \end{aligned}$

    $ \begin{aligned} \text { (b) }\quad \sqrt[a]{\dfrac{\sqrt[b]{x}}{\sqrt[c]{x}}} \cdot \sqrt[b]{\dfrac{\sqrt[c]{x}}{\sqrt[a]{x}}} \cdot \sqrt[c]{\dfrac{\sqrt[a]{x}}{\sqrt[b]{x}}} &= \sqrt[a]{\dfrac{x^{\frac{1}{b}}}{x^{\frac{1}{c}}}} \cdot \sqrt[b]{\dfrac{x^{\frac{1}{c}}}{x^{\frac{1}{a}}}} \cdot \sqrt[c]{\dfrac{x^{\frac{1}{a}}}{x^{\frac{1}{b}}}} \\\\ &= \dfrac{x^{\frac{1}{a b}}}{x^{\frac{1}{a c}}} \cdot \dfrac{x^{\frac{1}{b c}}}{x^{\frac{1}{b b}}} \cdot \dfrac{x^{\frac{1}{a c}}}{x^{\frac{1}{b c}}} \\\\ &=1 \end{aligned}$

    $\begin{aligned} \text { (c) }\quad \left(\dfrac{x^{m}-y^{m}}{x^{\frac{m}{2}}-y^{\frac{m}{2}}}-\frac{x^{m}-y^{m}}{x^{\frac{m}{2}}+y^{\frac{m}{2}}}\right)^{-2} &=\left(\dfrac{\left(x^{\frac{m}{2}}\right)^{2}-\left(y^{\frac{m}{2}}\right)^{2}}{x^{\frac{m}{2}}-y^{\frac{m}{2}}}-\dfrac{\left(x^{\frac{m}{2}}\right)^{2}-\left(y^{\frac{m}{2}}\right)^{2}}{x^{\frac{m}{2}}+y^{\frac{m}{2}}}\right)^{-2} \\\\ &=\left(\dfrac{\left(x^{\frac{m}{2}}+y^{\frac{m}{2}}\right)\left(x^{\frac{m}{2}}-y^{\frac{m}{2}}\right)}{x^{\frac{m}{2}}-y^{\frac{m}{2}}}-\dfrac{\left(x^{\frac{m}{2}}+y^{\frac{m}{2}}\right)\left(x^{\frac{m}{2}}-y^{\frac{m}{2}}\right)}{x^{\frac{m}{2}}+y^{\frac{m}{2}}}\right) \\\\ &=\left(x^{\frac{m}{2}}+y^{\frac{m}{2}}-x^{\frac{m}{2}}+y^{\frac{m}{2}}\right)^{-2} \\\\ &=\left(2 y^{\frac{m}{2}}\right)^{-2}=2^{-2} \cdot y^{-m}=\dfrac{1}{4 y^{m}} \end{aligned}$


Exercise 2.3

  1. Write the following in radical form.

  2. (a)  $(5)^{\frac{1}{2}}$  (b)  $(-9)^{\frac{1}{3}}$  (c)  $(2)^{-\frac{1}{2}}$

    (d)  $\left(-\dfrac{3}{4}\right)^{\frac{2}{5}}$ (e)  $\left(\dfrac{2}{7}\right)^{\frac{5}{2}}$


    (a)  $(5)^{\frac{1}{2}}=\sqrt{5}$

    (b)  $(-9)^{\frac{1}{3}}=\sqrt[3]{-9}$

    (c)  $(2)^{-\frac{1}{2}}=\dfrac{1}{2^{\frac{1}{2}}}=\dfrac{1}{\sqrt{2}}$

    (d)  $\left(-\dfrac{3}{4}\right)^{\frac{2}{5}}=\sqrt[5]{\left(-\dfrac{3}{4}\right)^{2}}=\sqrt[5]{\dfrac{9}{16}}$

    (e)  $\left(\dfrac{2}{7}\right)^{\frac{5}{2}}=\sqrt{\left(\dfrac{2}{7}\right)^{5}}=\sqrt{\dfrac{32}{16807}}$


  3. Write the following in fractional exponent form.

  4. (a)  $\sqrt[6]{c^{5}}$    (b)  $\sqrt[3]{-2}$

    (c)  $\sqrt[5]{a^{4} \sqrt[3]{b^{5}}}$ (d)  $\sqrt[4]{\left(\dfrac{3}{7}\right)^{3}}$


    (a)  $\sqrt[6]{c^{5}}=c^{\frac{5}{6}}$

    (b)  $\sqrt[3]{-2}=(-2)^{\frac{1}{3}}$

    (c)  $\sqrt[5]{a^{4} \sqrt[3]{b^{5}}}=\sqrt[5]{a^{4} \cdot b^{\frac{5}{3}}}=\left(a^{4} \cdot b^{\frac{5}{3}}\right)^{\frac{1}{5}}=a^{\frac{4}{5}} \cdot b^{\frac{1}{3}}$

    (d)  $\sqrt[4]{\left(\dfrac{3}{7}\right)^{3}}=\left(\dfrac{3}{7}\right)^{\frac{3}{4}}$


  5. Change the expression with the same radical and simplify the radicands.

  6. (a)  $6 \sqrt{2}$   (b)  $3 a \sqrt[3]{x}$  (c)  $2 \sqrt[5]{2}$

    (d)  $3 \sqrt[4]{\dfrac{1}{2}}$ (e)  $3 \sqrt{x^{3}}$


    (a)  $6 \sqrt{2}=\sqrt{6^{2} \cdot 2}=\sqrt{72}$

    (b)  $3 a \sqrt[3]{x}=\sqrt[3]{(3 a)^{3} x}=\sqrt[3]{27 a^{3} x}$

    (c)  $2 \sqrt[5]{2}=\sqrt[5]{2^{5} \cdot 2}=\sqrt[5]{64}$

    (d)  $3 \sqrt[4]{\dfrac{1}{2}}=\sqrt[4]{3^{4} \cdot \dfrac{1}{2}}=\sqrt[4]{\dfrac{81}{2}}$

    (c)  $3 \sqrt{x^{3}}=\sqrt{3^{2} \cdot x^{3}}=\sqrt{9 x^{3}}$


  7. Simplify.

  8. (a)  $\sqrt{32}$      (b)  $\sqrt[5]{-32}$

    (c)  $\sqrt[4]{\dfrac{81 x^{16}}{16 y^{4}}}$    (d)  $\sqrt[3]{\dfrac{81 x^{2}}{4 y}}$

    (e)  $\dfrac{9^{\frac{1}{2}}}{\sqrt[3]{27}}$      (f)  $\sqrt{\dfrac{2}{3}} \cdot \sqrt{\dfrac{75}{98}}$

    (g)  $\sqrt[3]{\dfrac{-216}{8 \times 10^{3}}}$  (h)  $\sqrt[n]{\dfrac{32}{2^{5+n}}}$


    (a)  $\sqrt{32}=\sqrt{16.2}=4 \sqrt{2}$

    (b)  $\sqrt[5]{-32}=\sqrt[5]{(-2)^{5}}=-2$

    (c)  $\sqrt[4]{\dfrac{81 x^{16}}{16 y^{4}}}=\sqrt[4]{\left(\dfrac{3 x^{4}}{2 y}\right)^{4}}=\dfrac{3 x^{4}}{2 y}$

    (d)  $\sqrt[3]{\dfrac{81 x^{2}}{4 y}}=\sqrt[3]{\dfrac{3^{3} \cdot 3 \cdot x^{2}}{2^{2} \cdot y}} \times \dfrac{\sqrt[3]{2 y^{2}}}{\sqrt[3]{2 y^{2}}}=\dfrac{3 \sqrt[3]{6 x^{2} y^{2}}}{2 y}$

    (e)  $\dfrac{9^{\frac{1}{2}}}{\sqrt[3]{27}}=\dfrac{\left(3^{2}\right)^{\frac{1}{2}}}{\sqrt[3]{3^{3}}}=\dfrac{3}{3}=1$

    (f)  $\sqrt{\dfrac{2}{3}} \cdot \sqrt{\dfrac{75}{98}}=\sqrt{\dfrac{2.75}{3.98}}=\sqrt{\dfrac{25}{49}}=\dfrac{5}{7}$

    (g)  $\sqrt[3]{\dfrac{-216}{8 \times 10^{3}}}=\sqrt[3]{\dfrac{(-6)^{3}}{2^{3} \cdot 10^{3}}}=\dfrac{-6}{2 \cdot 10}=\dfrac{-3}{10}$

    (h)  $\sqrt[n]{\dfrac{32}{2^{5+n}}}=\sqrt[n]{\dfrac{2^{5}}{2^{5+n}}}=\sqrt[n]{2^{-n}}=\left(2^{-n}\right)^{\frac{1}{n}}=2^{-1}=\dfrac{1}{2}$


  9. Rationalize the denominators.

  10. (a)  $\dfrac{4 \sqrt{35}}{3 \sqrt{7}}$    (b)  $\dfrac{20}{\sqrt{5}}$

    (c)  $\dfrac{18}{\sqrt[3]{2}} $     (d)  $\dfrac{\sqrt[3]{32}}{\sqrt[4]{27}}$

    (e)  $\dfrac{\sqrt[3]{36 a^{2}}}{\sqrt[3]{9 a}}$   (f)  $\dfrac{\sqrt[3]{2}}{\sqrt[6]{12}}$

    (g)  $\dfrac{1}{\sqrt[3]{x y^{2}}}$  (h)  $\sqrt[m]{\dfrac{2 x^{2} y^{3 m}}{9 x^{5} y^{4 m-1}}}$


    (a)  $\dfrac{4 \sqrt{35}}{3 \sqrt{7}}=\dfrac{4}{3}\times \sqrt{\dfrac {35}{7}}=\dfrac{4 \sqrt{5}}{3}$

    (b)  $\dfrac{20}{\sqrt{5}}=\dfrac{20}{\sqrt{5}} \times \dfrac{\sqrt{5}}{\sqrt{5}}=\dfrac{20 \sqrt{5}}{5}=4 \sqrt{5}$

    (c)  $\dfrac{18}{\sqrt[3]{2}}=\dfrac{18}{\sqrt[3]{2}} \times \dfrac{\sqrt[3]{2^{2}}}{\sqrt[3]{2^{2}}}=\dfrac{18 \sqrt[3]{4}}{2}=9 \sqrt[3]{4}$

    (d)  $\dfrac{18}{\sqrt[3]{2}} =\dfrac{18}{\sqrt[3]{2}} \times \dfrac{\sqrt[3]{2^{2}}}{\sqrt[3]{2^{2}}}=\dfrac{\sqrt[3]{2^{5}}}{\sqrt[4]{3^{3}}} \times \dfrac{\sqrt[4]{3}}{\sqrt[4]{3}}=\dfrac{2 \cdot \sqrt[3]{4} \cdot \sqrt[4]{3}}{3}$

    (e)  $\dfrac{\sqrt[3]{36 a^{2}}}{\sqrt[3]{9 a}}=\sqrt[3]{\dfrac{36 a^{2}}{9 a}}=\sqrt[3]{4 a}$

    (f)  $\dfrac{\sqrt[3]{2}}{\sqrt[6]{12}}=\dfrac{\sqrt[6]{4}}{\sqrt[6]{12}}=\dfrac{1}{\sqrt[6]{3}} \times \dfrac{\sqrt[6]{3^{5}}}{\sqrt[6]{3^{5}}}=\dfrac{\sqrt[6]{3^{5}}}{3}$

    (g)  $\dfrac{1}{\sqrt[3]{x y^{2}}} \times \dfrac{\sqrt[3]{x^{2} y}}{\sqrt[3]{x^{2} y}}=\dfrac{\sqrt[3]{x^{2} y}}{\sqrt[3]{(x y)^{3}}}=\dfrac{\sqrt[3]{x^{2} y}}{x y}$

    (h)  $\sqrt[m]{\dfrac{2 x^{2} y^{3 m}}{9 x^{5} y^{4 m-1}}}=\sqrt[m]{\dfrac{2}{9 x^{3} y^{m-1}}} \times \dfrac{\sqrt[m]{3^{m-2} \cdot x^{m-3} \cdot y}}{\sqrt[m]{3^{m-2} \cdot x^{m-3} \cdot y}}=\dfrac{\sqrt[m]{2 \cdot 3^{m-2} x^{m-3} y}}{3 x y}$


  11. Reduce the order as far as possible.

  12. (a)  $\sqrt[4]{25}$  (d)  $\sqrt[9]{8 y^{3}}$   (g)  $\sqrt[12]{64 a^{2} b^{6}}$

    (b)  $\sqrt[6]{4}$   (e)  $\sqrt[6]{27^{3}}$   (h)  $(72)^{\frac{3}{5}}$

    (c)  $\sqrt[6]{8}$   (f)  $\sqrt[8]{a^{2} b^{4}}$ (i)  $\sqrt[3]{768}$


    (a)  $\sqrt[4]{25}=\sqrt[4]{5^{2}}=\sqrt{5}$

    (b)  $\sqrt[6]{4}=\sqrt[6]{2^{2}}=\sqrt[3]{2}$

    (c)  $\sqrt[6]{8}=\sqrt[6]{2^{3}}=\sqrt{2}$

    (d)  $\sqrt[9]{8 y^{3}}=\sqrt[9]{(2 y)^{3}}=\sqrt[3]{2 y}$

    (e)  $\sqrt[6]{27^{3}}=\sqrt{27}$

    (f)  $\sqrt[8]{a^{2} b^{4}}=\sqrt[8]{\left(a b^{2}\right)^{2}}=\sqrt[4]{a b^{2}}$

    (g)  $\sqrt[12]{64 a^{2} b^{6}}=\sqrt[12]{\left(8 a b^{3}\right)^{2}}=\sqrt[6]{8 a b^{3}}$

    (h)  $(72)^{\frac{3}{5}}=\left(2^{3} \cdot 3^{2}\right)^{\frac{3}{5}}=\sqrt[5]{\left(2^{3} \cdot 3^{2}\right)^{3}}=\sqrt[5]{2^{9} \cdot 3^{6}}=2 \cdot 3 \sqrt[5]{2^{4} \cdot 3}=6 \sqrt[5]{48}$

    (i)  $\sqrt[3]{768}=\sqrt[3]{2^{8} \cdot 3}=2 \cdot 2 \sqrt[3]{2^{2} \cdot 3}=4 \sqrt[3]{12}$


  13. Find the simplified forms.

  14. (a)  $\sqrt{\dfrac{9}{50}}$    (c)  $\sqrt[4]{16}$

    (b)  $\sqrt[3]{\dfrac{-192}{49}}$   (d)  $2 \sqrt[3]{56}$


    (a)  $\sqrt{\dfrac{9}{50}}=\sqrt{\dfrac{3^{2}}{5^{2} \cdot 2}}=\dfrac{3}{5} \cdot \dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{3 \sqrt{2}}{10}$

    (b)  $\sqrt[3]{\dfrac{-192}{49}}=\sqrt[3]{\dfrac{-192}{7^{2}}} \times \dfrac{\sqrt[3]{7}}{\sqrt[3]{7}}=\dfrac{\sqrt[3]{(-4)^{3} \cdot 3 \cdot 7}}{7}$

    (c)  $\sqrt[4]{16}=\sqrt[4]{2^{4}}=2$

    (d)  $2 \sqrt[3]{56}=2 \sqrt[3]{2^{3} \cdot 7}=2 \cdot 2 \sqrt[3]{7}=4 \sqrt[3]{7}$


Exercise 2.4

  1. Simplify the following.

  2. (a)  $3 \sqrt{5}+7 \sqrt{5}$

    (b)  $\sqrt{75}-\sqrt{12}$

    (c)  $3 \cdot 3 \sqrt{3} \cdot 3 \sqrt{27}$

    (d)  $2 \sqrt{5} \cdot 3 \sqrt{2}$

    (e)  $(4-\sqrt{3})^{2}=16-8 \sqrt{3}+3$

    (f)  $(\sqrt{3}+2 \sqrt{2})(\sqrt{3}+\sqrt{2})$

    (g)  $(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6})(\sqrt{x}+1)(\sqrt{x}-1)$

    (h)  $\sqrt{75}-\frac{3}{4} \sqrt{48}-5 \sqrt{12}$

    (i)  $\sqrt{2 x^{2}}+5 \sqrt{32 x^{2}}-2 \sqrt{98 x^{2}}$

    (j)  $\sqrt{20 a^{3}}+a \sqrt{5 a}+\sqrt{80 a^{3}}$


    (a)  $3 \sqrt{5}+7 \sqrt{5}=10 \sqrt{5}$

    (b)  $\sqrt{75}-\sqrt{12}=5 \sqrt{3}-2 \sqrt{3}=3 \sqrt{3}$

    (c)  $3 \cdot 3 \sqrt{3} \cdot 3 \sqrt{27}=27 \sqrt{3 \cdot 27}=27 \cdot 9=243$

    (d)  $2 \sqrt{5} \cdot 3 \sqrt{2}=6 \sqrt{10}$

    (e)  $(4-\sqrt{3})^{2}=16-8 \sqrt{3}+3=19-8 \sqrt{3}$

    (f)  $(\sqrt{3}+2 \sqrt{2})(\sqrt{3}+\sqrt{2})=3+\sqrt{6}+2 \sqrt{6}+4=7+3 \sqrt{6}$

    (g)  $(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6})(\sqrt{x}+1)(\sqrt{x}-1)=\left(7^{2}-\sqrt{6}^{2}\right)\left(\sqrt{x}^{2}-1\right)=(7-6)(x-1)=x-1$

    (h)  $\sqrt{75}-\frac{3}{4} \sqrt{48}-5 \sqrt{12}=5 \sqrt{3}-\frac{3}{4} \cdot 4 \sqrt{3}-10 \sqrt{3}=-8 \sqrt{3}$

    (i)  $\sqrt{2 x^{2}}+5 \sqrt{32 x^{2}}-2 \sqrt{98 x^{2}}=\sqrt{2 x^{2}}+5 \cdot 4 \sqrt{2 x^{2}}-2 \cdot 7 \sqrt{2 x^{2}}=\sqrt{2 x^{2}}+20 \sqrt{2 x}-14 \sqrt{2 x}$

    (j)  $\sqrt{20 a^{3}}+a \sqrt{5 a}+\sqrt{80 a^{3}}=2 a \sqrt{5 a}+a \sqrt{5 a}+4 a \sqrt{5 a}=7 a \sqrt{5 a}$


  3. Rationalize the denominators and simplify.

  4. (a)  $\dfrac{2}{\sqrt{5}}$       (b)  $\dfrac{5}{2+\sqrt{3}}$

    (c)  $\dfrac{12}{\sqrt{5}-\sqrt{3}}$   (d)  $\dfrac{\sqrt{2}+1}{2 \sqrt{2}-1}$

    (e)  $\dfrac{\sqrt{7}+3 \sqrt{2}}{\sqrt{7}-\sqrt{2}}$  (f)  $\dfrac{\sqrt{17}-\sqrt{11}}{\sqrt{17}+\sqrt{11}}$

    (g)  $\dfrac{1}{2 \sqrt{2}-\sqrt{3}} $  (h)  $\dfrac{\sqrt{6}+1}{3-\sqrt{5}}$


    (a) $\dfrac{2}{\sqrt{5}}=\dfrac{2}{\sqrt{5}} \times \dfrac{\sqrt{5}}{\sqrt{5}}=\dfrac{2 \sqrt{5}}{5}$

    $\begin{aligned} \text { (b) }\quad \dfrac{5}{2+\sqrt{3}}&=\dfrac{5}{2+\sqrt{3}} \times \dfrac{2-\sqrt{3}}{2-\sqrt{3}}\\\\ &=\dfrac{5(2-\sqrt{3})}{2^{2}-\sqrt{3}^{2}}\\\\ &=\dfrac{10-5 \sqrt{3}}{4-3}\\\\ &=10-5 \sqrt{3} \end{aligned}$

    $\begin{aligned} \text { (c) }\quad \dfrac{12}{\sqrt{5}-\sqrt{3}}&=\dfrac{12}{\sqrt{5}-\sqrt{3}} \times \dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}\\\\ &=\dfrac{12(\sqrt{5}+\sqrt{3})}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}\\\\ &=\dfrac{12(\sqrt{5}+\sqrt{3})}{5-3}\\\\ &=6(\sqrt{5}+\sqrt{3}) \end{aligned}$

    $\begin{aligned} \text { (d) }\quad \dfrac{\sqrt{2}+1}{2 \sqrt{2}-1} &=\dfrac{\sqrt{2}+1}{2 \sqrt{2}-1} \times \dfrac{2 \sqrt{2}+1}{2 \sqrt{2}+1} \\\\ &=\dfrac{4+\sqrt{2}+2 \sqrt{2}+1}{(2 \sqrt{2})^{2}-1} \\\\ &=\dfrac{5+3 \sqrt{2}}{8-1} \\\\ &=\dfrac{5+3 \sqrt{2}}{7} \end{aligned}$

    $\begin{aligned} \text { (e) }\quad \dfrac{\sqrt{7}+3 \sqrt{2}}{\sqrt{7}-\sqrt{2}} &=\dfrac{\sqrt{7}+3 \sqrt{2}}{\sqrt{7}-\sqrt{2}} \times \dfrac{\sqrt{7}+\sqrt{2}}{\sqrt{7}+\sqrt{2}} \\\\ &=\dfrac{7+\sqrt{14}+3 \sqrt{14}+6}{(\sqrt{7})^{2}-(\sqrt{2})^{2}} \\\\ &=\dfrac{13+4 \sqrt{14}}{7-2} \\\\ &=\dfrac{13+4 \sqrt{14}}{5} \end{aligned}$

    $ \begin{aligned} \text { (f) }\quad \dfrac{\sqrt{17}-\sqrt{11}}{\sqrt{17}+\sqrt{11}} &=\dfrac{\sqrt{17}-\sqrt{11}}{\sqrt{17}+\sqrt{11}} \times \dfrac{\sqrt{17}-\sqrt{11}}{\sqrt{17}-\sqrt{11}} \\\\ &=\dfrac{17-\sqrt{17} \times \sqrt{11}-\sqrt{17} \times \sqrt{11}+11}{(\sqrt{17})^{2}-(\sqrt{11})^{2}} \\\\ &=\dfrac{28-2 \sqrt{187}}{17-11} \\\\ &=\dfrac{2(14-\sqrt{187})}{6} \\\\ &=\dfrac{14-\sqrt{187}}{3} \\\\ \end{aligned}$

    $ \begin{aligned} \text { (g) }\quad \dfrac{1}{2 \sqrt{2}-\sqrt{3}} &=\dfrac{1}{2 \sqrt{2}-\sqrt{3}} \times \dfrac{2 \sqrt{2}+\sqrt{3}}{2 \sqrt{2}+\sqrt{3}} \\\\ &=\dfrac{2 \sqrt{2}+\sqrt{3}}{(2 \sqrt{2})^{2}-(\sqrt{3})^{2}} \\\\ &=\dfrac{2 \sqrt{2}+\sqrt{3}}{8-3} \\\\ &=\dfrac{2 \sqrt{2}+\sqrt{3}}{5} \end{aligned}$

    $\begin{aligned} \text { (h) }\quad\dfrac{\sqrt{6}+1}{3-\sqrt{5}} &=\dfrac{\sqrt{6}+1}{3-\sqrt{5}} \times \dfrac{3+\sqrt{5}}{3+\sqrt{5}} \\\\ &=\dfrac{3 \sqrt{6}+\sqrt{30}+\sqrt{5}+3}{3^{2}-(\sqrt{5})^{2}} \\\\ &=\dfrac{3 \sqrt{6}+\sqrt{30}+\sqrt{5}+3}{9-5} \\\\ &=\dfrac{3 \sqrt{6}+\sqrt{30}+\sqrt{5}+3}{4} \end{aligned}$


  5. Write as a single fraction.

  6. (a)  $\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}$

    (b)  $\dfrac{2}{\sqrt{7}+\sqrt{2}}+\dfrac{1}{\sqrt{7}-\sqrt{2}}$

    (c)  $\dfrac{1}{3+\sqrt{3}}+\dfrac{1}{\sqrt{3}-3}+\frac{1}{\sqrt{3}}$

    (d)  $\dfrac{7+\sqrt{5}}{7-\sqrt{5}}+\dfrac{\sqrt{11}-3}{\sqrt{11}+3}$

    (e)  $\dfrac{3+2 \sqrt{2}}{(\sqrt{3}-1)^{2}}$

    (f)  $\sqrt{\dfrac{x+1}{x-1}}+\sqrt{\dfrac{x-1}{x+1}}+\sqrt{\dfrac{1}{x^{2}-1}}$

    (g)  $\sqrt{\dfrac{\sqrt[5]{32}+\sqrt{4}}{2^{-2}-2^{-3}}}$


    $\begin{aligned} \text { (a) }\quad \dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}&=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{(\sqrt{3}+1)(\sqrt{3}-1)}\\\\ &=\dfrac{2 \sqrt{3}}{(\sqrt{3})^{2}-1^{2}}\\\\ &=\dfrac{2 \sqrt{3}}{3-1}\\\\ &=\dfrac{2 \sqrt{3}}{2}\\\\ &=\sqrt{3} \end{aligned}$

    $\begin{aligned} \text { (b) }\quad \dfrac{2}{\sqrt{7}+\sqrt{2}}+\dfrac{1}{\sqrt{7}-\sqrt{2}}&=\dfrac{2(\sqrt{7}-\sqrt{2})+1(\sqrt{7}+\sqrt{2})}{(\sqrt{7})^{2}-(\sqrt{2})^{2}}\\\\ &=\dfrac{2 \sqrt{7}-2 \sqrt{2}+\sqrt{7}+\sqrt{2}}{7-2}\\\\ &=\dfrac{3 \sqrt{7}-\sqrt{2}}{5} \end{aligned}$

    $\begin{aligned} \text { (c) }\quad \dfrac{1}{3+\sqrt{3}}+\dfrac{1}{\sqrt{3}-3}+\frac{1}{\sqrt{3}} &=\dfrac{\sqrt{3}-3+3+\sqrt{3}}{(\sqrt{3}+3)(\sqrt{3}-3)}+\dfrac{1}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\\\ &=\dfrac{2 \sqrt{3}}{(\sqrt{3})^{2}-3^{2}}+\dfrac{\sqrt{3}}{3} \\\\ &=\dfrac{2 \sqrt{3}}{3-9}+\dfrac{\sqrt{3}}{3} \\\\ &=\dfrac{2 \sqrt{3}}{-6}+\dfrac{\sqrt{3}}{3} \\\\ &=\dfrac{-\sqrt{3}}{3}+\dfrac{\sqrt{3}}{3} \\\\ &=0 \end{aligned}$

    $ \begin{aligned} \text { (d) }\quad \dfrac{7+\sqrt{5}}{7-\sqrt{5}}+\dfrac{\sqrt{11}-3}{\sqrt{11}+3} &=\dfrac{7+\sqrt{5}}{7-\sqrt{5}} \times \dfrac{7+\sqrt{5}}{7+\sqrt{5}}+\dfrac{\sqrt{11}-3}{\sqrt{11}+3} \times \dfrac{\sqrt{11}-3}{\sqrt{11}-3} \\\\ &=\dfrac{49+7 \sqrt{5}+7 \sqrt{5}+5}{7^{2}-(\sqrt{5})^{2}}+\dfrac{11-3 \sqrt{11}-3 \sqrt{11}+9}{(\sqrt{11})^{2}-3^{2}} \\\\ &=\dfrac{54+14 \sqrt{5}}{49-5}+\dfrac{20-6 \sqrt{11}}{11-9} \\\\ &=\dfrac{54+14 \sqrt{5}}{44}+\dfrac{20-6 \sqrt{11}}{2} \\\\ &=\dfrac{2(27+7 \sqrt{5})}{44}+\dfrac{20-6 \sqrt{11}}{2} \\\\ &=\dfrac{27+7 \sqrt{5}}{22}+\dfrac{20-6 \sqrt{11}}{2} \\\\ &=\dfrac{27+7 \sqrt{5}+220-66 \sqrt{11}}{22} \\\\ &=\dfrac{247+7 \sqrt{5}-66 \sqrt{11}}{22} \end{aligned}$

    $\begin{aligned} \text { (e) }\quad \dfrac{3+2 \sqrt{2}}{(\sqrt{3}-1)^{2}} &=\dfrac{3+2 \sqrt{2}}{3-2 \sqrt{3}+1} \\\\ &=\dfrac{3+2 \sqrt{2}}{4-2 \sqrt{3}} \times \dfrac{4+2 \sqrt{3}}{4+2 \sqrt{3}} \\\\ &=\dfrac{12+6 \sqrt{3}+8 \sqrt{2}+4 \sqrt{6}}{4^{2}-(2 \sqrt{3})^{2}} \\\\ &=\dfrac{12+6 \sqrt{3}+8 \sqrt{2}+4 \sqrt{6}}{16-12} \\\\ &=\dfrac{2(6+3 \sqrt{3}+4 \sqrt{2}+2 \sqrt{6})}{4} \\\\ &=\dfrac{6+4 \sqrt{2}+3 \sqrt{3}+2 \sqrt{6}}{2} \end{aligned}$

    $\begin{aligned} \text { (f) }\quad \sqrt{\dfrac{x+1}{x-1}}+\sqrt{\dfrac{x-1}{x+1}}+\sqrt{\dfrac{1}{x^{2}-1}}&=\dfrac{\sqrt{x+1}}{\sqrt{x-1}}+\dfrac{\sqrt{x-1}}{\sqrt{x+1}}-\dfrac{1}{\sqrt{x^{2}-1}}\\\\ &=\dfrac{\sqrt{x+1}}{\sqrt{x-1}} \times \dfrac{\sqrt{x-1}}{\sqrt{x-1}}+\dfrac{\sqrt{x-1}}{\sqrt{x+1}} \times \dfrac{\sqrt{x+1}}{\sqrt{x+1}}-\dfrac{1}{\sqrt{x^{2}-1}} \times \dfrac{\sqrt{x^{2}-1}}{\sqrt{x^{2}-1}}\\\\ &=\dfrac{\sqrt{x^{2}-1}}{x-1}+\dfrac{\sqrt{x^{2}-1}}{x+1}-\dfrac{\sqrt{x^{2}-1}}{x^{2}-1}\\\\ &=\dfrac{(x+1) \sqrt{x^{2}-1}+(x-1) \sqrt{x^{2}-1}-\sqrt{x^{2}-1}}{x^{2}-1}\\\\ &=\dfrac{x \sqrt{x^{2}-1}+\sqrt{x^{2}-1}+x \sqrt{x^{2}-1}-\sqrt{x^{2}-1}-\sqrt{x^{2}-1}}{x^{2}-1}\\\\ &=\dfrac{2 x \sqrt{x^{2}-1}-\sqrt{x^{2}-1}}{x^{2}-1}\\\\ &=\dfrac{(2 x-1) \sqrt{x^{2}-1}}{x^{2}-1} \end{aligned}$

    $\begin{aligned} \text { (g) }\quad \sqrt{\dfrac{\sqrt[5]{32}+\sqrt{4}}{2^{-2}-2^{-3}}}&=\sqrt{\dfrac{\sqrt[5]{2^{5}}+\sqrt{4}}{\dfrac{1}{2^{2}}-\dfrac{1}{2^{3}}}}\\\\ &=\sqrt{\dfrac{2+2}{\dfrac{1}{4}-\dfrac{1}{8}}}\\\\ &=\sqrt{\dfrac{4}{\dfrac{2-1}{8}}}\\\\ &=\sqrt{32}\\\\ &=\sqrt{16 \times 2}\\\\ &=4 \sqrt{2} \end{aligned}$


Exercise 2.5

Solve the following equations.

1.  $3^{2 x-3} =27^{2 x}$

2.  $5^{x^{2}-9}=1$

3.  $5^{x+1} =\dfrac{1}{625}$

4.  $\left(\dfrac{1}{2}\right)^{x} =64$

5.  $ 2^{3 x} \cdot 4^{x+1}=128 $

6.  $3^{x+1} \cdot 9^{2-x}=\dfrac{1}{27}$

7.  $\dfrac{27^{2 x}}{3^{5-x}}= \dfrac{3^{2 x+1}}{9^{x+3}}$

8.  $8^{x-1} =\left(\dfrac{1}{32}\right)^{x+1}$

9.  $10^{-x} =0.000001$

10.  $4^{x}+4^{x+1}=20$

11.  $4 \cdot 2^{2 x}+3 \cdot 2^{x}-1=0$


$\begin{aligned} \text { 1. }\quad 3^{2 x-3} &=27^{2 x} \\\\ 3^{2 x-3} &=\left(3^{3}\right)^{2 x} \\\\ 3^{2 x-3} &=3^{6 x} \\\\ 2 x-3 &=6 x \\\\ 4 x &=-3 \\\\ x &=-\dfrac{3}{4} \end{aligned}$

$\begin{aligned} \text { 2. }\quad\quad\quad\quad\quad5^{x^{2}-9}&=1 \\\\ 5^{x^{2}-9}&=5^{0} \\\\ x^{2}-9&=0 \\\\ (x+3)(x-3)&=0\\\\ x+3=0 \text{ (or) } x-3&=0\\\\ x=-3\text{ (or) } x&=3 \end{aligned}$

$\begin{aligned} \text { 3. }\quad5^{x+1} &=\dfrac{1}{625} \\\\ 5^{x+1} &=\dfrac{1}{5^{4}} \\\\ 5^{x+1} &=5^{-4} \\\\ x+1 &=4 \\\\ x &=3 \end{aligned}$

$\begin{aligned} \text { 4. }\quad\left(\dfrac{1}{2}\right)^{x} &=64 \\\\ 2^{-x} &=2^{6} \\\\ -x &=6 \\\\ x &=-6 \end{aligned}$

$\begin{aligned} \text { 5. }\quad 2^{3 x} \cdot 4^{x+1}&=128 \\\\ 2^{3 x} \cdot\left(2^{2}\right)^{x+1}&= 2^{7} \\\\ 2^{3 x} \cdot 2^{2 x+2} & =2^{7} \\\\ 3 x+2 x+2 &= 7 \\\\ 5 x+2 & =7 \\ 5 x & =5 \\ x & =1 \end{aligned}$

$\begin{aligned} \text { 6. }\quad 3^{x+1} \cdot 9^{2-x}&=\dfrac{1}{27}\\\\ 3^{x+1} \cdot\left(3^{2}\right)^{2-x}&=\dfrac{1}{3^{3}}\\\\ 3^{x+1} \cdot 3^{4-2 x}&=3^{-3}\\\\ 3^{x+1+4-2 x} &=3^{-3} \\\\ -x+5 &=-3 \\\\ x &=8 \end{aligned}$

$\begin{aligned} \text { 7. }\quad\dfrac{27^{2 x}}{3^{5-x}}&= \dfrac{3^{2 x+1}}{9^{x+3}} \\\\ \dfrac{\left(3^{3}\right)^{2 x}}{3^{5-x}}&=\dfrac{3^{2 x+1}}{\left(3^{2}\right)^{x+3}} \\\\ \dfrac{3^{6 x}}{3^{5-x}} &=\dfrac{3^{2 x+1}}{3^{2 x+6}} \\\\ 3^{6 x-5+x} &=3^{2 x+1-2 x-6} \\\\ 3^{7 x-5} &=3^{-5} \\\\ 7 x-5 &=-5 \\\\ 7 x &=0 \\\\ x &=0 \end{aligned}$

$\begin{aligned} \text { 8. }\quad 8^{x-1} &=\left(\dfrac{1}{32}\right)^{x+1} \\\\ \left(2^{3}\right)^{x-1} &=\left(\dfrac{1}{2^{5}}\right)^{x+1} \\\\ 2^{3 x-3} &=\left(2^{-5}\right)^{x+1} \\\\ 2^{3 x-3} &=2^{-5 x-5} \\\\ 3 x-3 &=-5 x-5 \\\\ 8 x &=-2 \\\\ x &=-\dfrac{2}{8} \\\\ x &=-\dfrac{1}{4} \end{aligned}$

$\begin{aligned} \text { 9. }\quad 10^{-x} &=0.000001 \\\\ 10^{-x} &=10^{-6} \\\\ -x &=-6 \\\\ x &=6 \end{aligned}$

$\begin{aligned} \text { 10. }\quad 4^{x}+4^{x+1}&=20\\\\ 4^{x}+4^{x} \cdot 4&=20\\\\ 4^{x}(1+4)&=20\\\\ 4^{x}&=\dfrac{20}{5}\\\\ 4^{x}&=4\\\\ x&=1 \end{aligned}$

$\begin{aligned} \text { 11. }\quad 4 \cdot 2^{2 x}+3 \cdot 2^{x}-1&=0 \\\\ 4\left(2^{x}\right)^{2}+3 \cdot 2^{x}-1&=0 \\\\ \end{aligned}$

$\begin{aligned} \text { Let } 2^{x}&=a \\\\ 4 a^{2}+3 a-1&=0 \\\\ (4 a-1)(a+1)&=0 \\\\ 4 a-1=0 \text { (or) } a+1&=0 \\\\ a=\dfrac{1}{4} \text { (or) } a&=-1 \\\\ 2^{x}=\dfrac{1}{2^{2}} \text { (or) } 2^{x}&=-1 \\\\ 2^{x}=2^{-2} \text { (or) }& \text { impossible } \end{aligned}$

  $\therefore x=-2$