Logarithms

Chapter 3

Logarithms

In the seventeenth century, a Scottish mathematician John Napier developed method for efficiently performing calculations with larger numbers. He found a method for finding the product of two numbers by adding two corresponding numbers, which is called logarithms.The Richter scale is a logarithmic scale use to measure the magnitude of an earthquate.pH is another logarithmic scale used to describe how acidic or basic an aqueous solution is.

(၁၇)ရာစုတွင်စကော့တလန်သင်္ချာပညာရှင် John Napier သည်အလွန်ကြီးသောကိန်းများကိုထိရောက်စွာကိုင်တွယ်နိုင်မည့်နည်းလမ်းကိုတီထွင်ခဲ့သည်။ကိန်းနှစ်လုံးမြှောက်လဒ်ရှာဖွေရန် logarithm ဟုခေါ်သောသက်ဆိုင်ရာကိန်းနှစ်လုံးပေါင်းထည့်သည့်နည်းလမ်းတစ်ခုရှာဖွေတွေ့ရှိခဲ့သည်။ Richter scale သည်မြေငလျင်တစ်ခု၏ပမာဏကိုတိုင်းတာရန်သုံးသော Logarithmic scale တစ်ခုဖြစ်သည်။ pH scale သည် aqueous solution တစ်ခုတွင်အက်ဆစ်ဓာတ်(သို့မဟုတ်)ဘေ့စ်ဓာတ်မည်မျှပါဝင်ကြောင်းဖော်ပြရာတွင်သုံးသည့်အခြားသော logarithmic scale တစ်ခုဖြစ်သည်။

သိပ္ပံပညာရှင်များသည်အလွန်ကြီးသော(သို့မဟုတ်)အလွန်သေးသောကိန်းဂဏန်းများ(ဉပမာအားဖြင့် the speed of light in vacuum is$299,792,458$metres per second and the mass of an electron is $0.000000000000000000000000000000910938$ kilograms)ကိုကိုင်တွယ်ဖြေရှင်းရာတွင် (scientific notation) ဟုခေါ်သောပို၍ကျစ်လစ်သောကိန်းပုံစံကိုတီထွင်အသုံးပြုလာကြသည်။

3.1 Scientific Notation

Definition A positive number is in scientific notation when it is wtitten in the form $a\times10^{n}$, where $n$ is an integer and $a$ is a real number satisfying the inequality $1≤a<10$. အပေါင်းကိန်းတစ်ခုကို $a\times10^{n}$ ပုံစံဖြင့်ရေးထားလျှင်ယင်းပုံစံကို scientific notation (standard form) ဟုခေါ်၏။ $n$ သည်ကိန်းပြည့်တစ်ခုဖြစ်ပြီးကိန်းစစ် $a$ သည် $1≤a<10$ ကိုပြေလည်စေသောကိန်းဖြစ်ရမည်။ ပေးထားသောကိန်းတစ်ခုကို scientific notation သို့ပြောင်းလိုလျှင် ဒသမသင်္ကေတကိုပထမဦးဆုံးသုညမဟုတ်သောဂဏန်း ($1$ နှင့် $10$ ကြား)၏နောက်တွင်ထားပါ။ ဒသမကိုမူလနေရာမှမိမိရွှေ့လိုသောနေရာသို့နေရာအရေအတွက်ရေတွက်ပါ။ဤသို့နေရာအ‌ရေအတွက်ရေတွက်ခြင်းသည် $10$ ၏ထပ်ညွှန်းပမာဏကိုရှာဖွေခြင်းဖြစ်သည်။ ဒသမကိုလက်ဝဲဘက်သို့ရွှေ့လျှင် $10$ ၏ထပ်ညွှန်းသည်အပေါင်းဖြစ်ပြီး၊လက်ယာဘက်သို့ရွှေ့လျှင် $10$ ၏ထပ်ညွှန်းသည်အနုတ်ဖြစ်သည်။ In scientific notation, The speed of light $= 2.99792458\times10^{8}$ $ms^{-1}$ The mass of an electron $= 9.10938\times10^{-31}$ $kg$

Significant Figures

Definition An approximate number wtitten in scientific notation $a\times10^{n}$ indicates an accuracy to the number of digits in "$a$".The figures or digits in "$a$" are called significant figures. အရာရောက်ဂဏန်းများအကြောင်းအောက်ပါနေရာတွင်လေ့လာပါ။ ဒီနေရာကိုနှိပ်၍ဖတ်ပါ

Calculations Involving Measurement

($1$)  In addition and subtraction, the result should be rounded so that it has the same number of digits after the decimal as the measurement with the least number of digits to the right of the decimal. ဒသမကိန်းနှစ်လုံးပေါင်းခြင်း(သို့မဟုတ်)နုတ်ခြင်းမှရသောရလဒ်ကိန်းကိုအနီးဆုံးအမှန်တန်ဖိုးယူရာတွင်ကိန်းနှစ်လုံးအနက်ဒသမနောက်ရှိနည်းသောကိန်းလုံးအရေအတွက်ကိုသာယူသည်။ Example $\begin{aligned} 100.1+54.52147&=154.62147\\\\ &\approx154.6 \end{aligned}$ $100.1$  (ဒသမနောက်တွင်ကိန်းတစ်လုံး) $54.52147$  (ဒသမနောက်တွင်ကိန်းငါးလုံး) $1$ လုံး < $5$ လုံးဖြစ်သောကြောင့်အဖြေကိုဒသမတစ်နေရာသာယူသည်။ ပိုမိုထင်ရှားစေရန် Example 3 (a), (b) တို့တွင်ထပ်မံလေ့လာကြည့်ပါ။ ($2$)  In multiplication and division, the result should have the same number of significant figures as the measurement with the least. ဒသမကိန်းနှစ်လုံးမြှောက်ခြင်း(သို့မဟုတ်)စားခြင်းမှရသောရလဒ်ကိန်းကိုအနီးဆုံးအမှန်တန်ဖိုးယူရာတွင်ကိန်းနှစ်လုံးအနက်နည်းသောအရာရောက်ဂဏန်းအရေအတွက်ကိုသာယူသည်။ Example $\begin{aligned} 5.01\times45.0536&=225.718536\\\\ &\approx226 \end{aligned}$ $5.01$ (အရာရောက်ဂဏန်းသုံးလုံး) $45.0536$  (အရာရောက်ဂဏန်းခြောက်လုံး) $3$ လုံး<$6$ လုံးဖြစ်သောကြောင့်အဖြေကိုအရာရောက်ဂဏန်းသုံးလုံးသာယူသည်။ ပိုမိုထင်ရှားစေရန် Example 3 (c), (d), (e), (f) နှင့် Example 4 တို့တွင်ထပ်မံလေ့လာကြည့်ပါ။

Example 1.

Write the following numbers in scientific notation.

(a) $14,753$  (b) $0.00632$  (c) $0.23$

(d) $0.00000912$  (e) $1,000,000$

(a)  $147,53=1.4753 \times 10^{4}$ [Decimal point is moved four places to the left.] (b)  $0.00632=6.32 \times 10^{-3}$ [Decimal point is moved three places to the right.] (c)  $0.23=2.3 \times 10^{-1}$ [Decimal point is moved one place to the right.] (d)  $0.00000912=9.12 \times 10^{-6}$ [Decimal point is moved six places to the right.] (e)  $1,000,000=1 \times 10^{6}$ [Decimal point is moved six places to the left.]

Example 2.

Express the following numbers in ordinary decimal form.

(a) $7.354\times10^{5}$   (b) $3.2\times10^{-1}$

(a) $7.354 \times 10^{5}=735400$ (b) $3.2 \times 10^{-1}=0.32$

Example 3.

Evaluate each of the followings and express the results in scientific notation:

(a) $4.215\times10^{-2}+3.2\times10^{-4}$ [Addition]

(b) $8.97\times10^{4}-2.62\times10^{3}$ [Subtraction]

(c) ($6.73\times10^{-5}$)($2.91\times10^{2}$) [Multiplication]

(d) $\dfrac{6.4\times10^{6}}{1.92\times10^{2}}$ [Division]

(e) ($6.5\times10^{-3})^{2}$ [Power]

(f) $\sqrt{3.6\times10^{5}}$ [Root]

$\begin{aligned} \text{ (a) }&\quad 4.5 \times 10^{-2}+3.2 \times 10^{-4}\\\\ &=4.215 \times 10^{-2}+0.032 \times 10^{-2} \\\\ &=(4.215+0.032) \times 10^{-2} \\\\ &=4.247 \times 10^{-2} \end{aligned}$ $ \begin{aligned} \text { (b) }&\quad 8.97 \times 10^{4}-2.62 \times 10^{3} \\\\ &= 8.97 \times 10^{4}-0.262 \times 10^{4} \\\\ &=\left(8.97-0.262\right)\times 10^{4}\\\\ &= 8.708 \times 10^{4} \\\\ &\approx 8.71 \times 10^{4} \end{aligned}$ $\begin{aligned} \text { (c) } &\quad\left(6.73 \times 10^{-5}\right)\left(2.91 \times 10^{2}\right) \\\\ &= 19.5843 \times 10^{-3} \\\\ &= 1.95843 \times 10^{-2} \\\\ &\approx 1.96 \times 10^{-2} \end{aligned}$ $\begin{aligned} \text { (d) }\quad \dfrac{6.4 \times 10^{6}}{1.92 \times 10^{2}}&=\dfrac{640}{192} \times 10^{4}\\\\ &=3.333 \times 10^{4}\\\\ &\approx 3.3 \times 10^{4} \end{aligned}$ $\begin{aligned} \text { (e) }\quad\left(6.5 \times 10^{-3}\right)^{2}&=42.25 \times 10^{-6}\\\\ &=4.225 \times 10^{-5} \\\\ &\approx 4.2 \times 10^{-5} \end{aligned}$ $\begin{aligned} \text { (f) }\quad \sqrt{3.6 \times 10^{5}}&=\sqrt{36 \times 10^{4}}\\\\ &=6.0 \times 10^{2} \end{aligned}$

Example 4.

Evaluate $\dfrac{2,750,000\times0.015}{750}$ by transforming each number to scientific notation.

$\begin{aligned} \dfrac{2,750,000 \times 0.015}{750}&=\dfrac{\left(2.75 \times 10^{6}\right)\left(1.5 \times 10^{-2}\right)}{7.5 \times 10^{2}}\\\\ &=0.55 \times 10^{2}\\\\ &=55 \end{aligned}$

3.2 Definition of the Logarithm

Exponential equation $2^{x}=8$ တွင် $x$ ၏တန်ဖိုး $3$ ဖြစ်ကြောင်းလွယ်ကူစွာသိနိုင်သည်။သို့သော် $2^{x}=\dfrac{1}{3}$ နှင့် $2.718^{x}=5$ ကဲ့သို့သောညီမျှခြင်းများတွင် $x$ ၏တန်ဖိုးကိုအလွယ်တကူမသိနိုင်ပေ။ထို့ကြောင့်ကိန်းစစ်များ၏အခြေခံဂုဏ်သတ္တိများသည် logarithm အဓိပ္ပါယ်သတ်မှတ်ရာတွင်အရေးကြီးသောအခန်းကဏ္ဍမှပါဝင်လာသည်။ Given a positive number $N$, and a positive number $b$ other than $1$, the equation $N=b^{x}$ has exactly one solution for $x$. အပေါင်းကိန်းတစ်ခု $N$ နှင့် $1$ မဟုတ်သောအပေါင်းကိန်းတစ်ခု $b$ ပေးထားလျှင် $N=b^{x}$ ညီမျှခြင်းတွင် $x$ အတွက်အတိအကျအဖြေတစ်ခုသာရှိသည်။ Definition Let $N$ and $b$ be positive real numbers, with $b\neq1$. Then the logarithm of $N$ (with respect) to the base $b$ is the exponent by which $b$ is the exponent by which $b$ must be raised to yield $N$, and is denoted by $\log_{b}N$. $N$ နှင့် $b\neq1$ တို့သည်အပေါင်းကိန်းများဖြစ်ပါစေ။ $x=\log_{b}N$ သည် $b^{x}=N$ ၏အဖြေဖြစ်သည်။ $b^{x}=N$ နှင့် $x=\log_{b}N$ တို့သည်အတူတူပင်ဖြစ်သည်။တနည်းအားဖြင့် $b^{x}=N$ ⇔ $x=\log_{b}N$ $10^{5}=100,000$ ⇔ $5=\log_{10}100,000$ The followings are immediate consequences of the definition of logarithm: အောက်ဖော်ပြပါတို့သည် logarithm အဓိပ္ပါယ်သတ်မှတ်ချက်နှင့်ဆက်စပ်သောအကျိုးဆက်များဖြစ်သည်။ $L1$. $N=b^{\log_{b}N}$ $L2$. $x=\log_{b}b^{x}$ $L3$. $\log_{b}b=1$ $L4$. $\log_{b}1=0$

Example 5.

Write each of the following in logarithmic form.

(a) $10^{-2}=0.01$ (b) $4^{\frac{1}{2}}=2$ (c) $7^{2}=49$

(a)  $\log _{10} 0.01=-2$ (b)  $\log _{4} 2=\dfrac{1}{2}$ (c)  $\log _{7} 49=2$

Example 6.

Express each of the following in exponential form.

(a) $\log_{2}8=3$    (b) $\log_{10}1=0$

(c) $\log_{5}\left(\dfrac{1}{\sqrt{5}}\right)=-\dfrac{1}{2}$

(a)  $2^{3}=8$ (b)  $10^{0}=1$ (c)  $5^{-\frac{1}{2}}=\frac{1}{5^{\frac{1}{2}}}=\frac{1}{\sqrt{5}}$

Example 7.

Find the value of each logarithm.

(a) $\log_{5}25$ (b) $\log_{2}16\sqrt{2}$ (c) $\log_{\frac{1}{2}}8$

(a)  $\log _{5} 25=\log _{5} 5^{2}=2$ (b)  $\log _{2} 16 \sqrt{2}=\log _{2} 2^{4} \cdot 2^{\frac{1}{2}}=\log _{2} 2^{\frac{9}{2}}=\frac{9}{2}$ (c)  $\log _{\frac{1}{2}} 8=\log _{\frac{1}{2}} 2^{3}=\log _{\frac{1}{2}}\left(\frac{1}{2}\right)^{-3}=-3$

Example 8.

Evaluate each expression.

(a) $3^{\log _{3} 7}+\log _{5} 125$  (b) $\log _{7} 7^{9}-\log _{3} \dfrac{1}{9}$

(c) $\log _{5}\left(\log _{2}\left(\log _{3} 9\right)\right)$  (d) $10^{2+\log _{10} 5}$

(a)  $3^{\log _{3} 7}+\log _{5} 125=7+\log _{5} 5^{3}=7+3=10$ (b)  $\log _{7} 7^{9}-\log _{3} \dfrac{1}{9}=9-\log _{3} 3^{-2}=9-(-2)=11$ $\begin{aligned} \text{ (c) }\quad \log _{5}\left(\log _{2}\left(\log _{3} 9\right)\right)&=\log _{5}\left(\log _{2}\left(\log _{3} 3^{2}\right)\right)\\\\ &=\log _{5}\left(\log _{2} 2\right) \\\\ &=\log _{5} 1 \\\\ &=0 \end{aligned}$ (d)  $10^{2+\log _{10} 5}=10^{2} \cdot 10^{\log _{10^{5}}}=100 \cdot 5=500$

Example 9.

(a) Given that $10^{0.3010}=2$, find the value of $\log_{10}16$.

(b) Solve $\log_{3}(x^{2}-1)=2$.

(a) Since $10^{0.3010}=2$, we have $\log _{10} 2=0.3010$ $\begin{aligned} \log _{10} 16&=\log _{10} 2^{4}\\\\ &=4 \log _{10} 2\\\\ &=4(0.3010)\\\\ &=1.204 \end{aligned}$ $\begin{aligned} \text{ (b) }\quad Since\log _{3}\left(x^{2}-1\right)&=2,\\\\ \text{ we have } x^{2}-1&=3^{2}\\\\ x^{2} &=10 \\\\ x &=\pm \sqrt{10} \end{aligned}$

3.3 Properties of Logarithm

Theorem 1 If $M$, $N$, $B$ are positive real numbers, $b\neq1$ and $p$ is any real number, then $L5$. $\log_{b}(MN)=\log_{b}M+\log_{b}N$ $L6$. $log_{b}N^{p}=p\log_{b}N$ $L7$. $\log_{b}\left(\dfrac{M}{N}\right)=\log_{b}M-\log_{b}N$ Proof: $L5$.  $M=b^{\log_{b}M}$, $N=b^{\log_{b}N}$, $\begin{aligned} M N&= b^{\log_{b}M}b^{\log_{b}N}\\\\ &=b^{\log_{b}N+\log_{b}N}\\\\ \log_{b}(MN)&=\log_{b}M+\log_{b}N \end{aligned}$ $L6$.  $N=b^{\log_{b}N}$, $\begin{aligned} N^{p}&=(b^{\log_{b}N})^{p}\\\\ &=b^{p\log_{b}N}\\\\ log_{b}N^{p}&=p\log_{b}N \end{aligned}$ $\begin{aligned} L7. \log_{b}\left(\dfrac{M}{N}\right)&=\log_{b}(M N^{-1})\\\\ &=\log_{b}M+\log_{b}N^{-1}\\\\ &=\log_{b}M-\log_{b}N \end{aligned}$

Basic properties of exponents and logarithms can be summarized as follow.

ထပ်ညွှန်း၏အခြေခံဂုဏ်သတ္တိများနှင့် logarithm ၏အခြေခံဂုဏ်သတ္တိများကိုယှဉ်တွဲလေ့လာနိုင်သည်။

Properties	For exponents	For Logarithms
One-to-one Property	If $b^{x}=b^{y}$, then $x=y$	If $\log_{b}M=\log_{b}N$, then $M=N$.
Product Property	$b^{x}.b^{y}=b^{x +y}$	$\log_{b}(M N)=\log_{b}M+\log_{b}N$
Quotient Property	$\dfrac{b^{x}}{b^{y}}=b^{x -y}$	$\log_{b}\dfrac{M}{N}=\log_{b}M-log_{b}N$
Power Property	$(b^{x})^{y}=b^{x y}$	$\log_{b}N^{p}=p\log_{b}N$

Example 10.

If $p=\log_{b}2$, $q=\log_{b}3$ and $r=\log_{b}5$, write $\log_{b}\dfrac{5\sqrt{2}}{2}$ in terms of $p$, $q$ and $r$.

$\begin{aligned} p=\log _{b} 2 &, q=\log _{b} 3 \text { and } r=\log _{b} 5 \\\\ \log _{b} \dfrac{5 \sqrt{3}}{2} &=\log _{b} 5+\log _{b} \sqrt{3}-\log _{b} 2 \\\\ &=r+\log _{b} 3^{\frac{1}{2}}-p \\\\ &=r+\dfrac{1}{2} \log _{b} 3-p \\\\ &=r+\dfrac{1}{2} q-p \end{aligned}$

Example 11.

Using $\log_{2}3=1.5850$, find the values of

(a) $\log_{2}24$    (b) $\log_{2}0.75$

$\log _{2} 3=1.5850$ (a) $\log _{2} 24=\log _{2}\left(2^{3} \times 3\right)=\log _{2} 2^{3}+\log _{2} 3=3+1.5850=4.5850$ (b) $\log _{2} 0.75=\log _{2}\left(\frac{3}{4}\right)=\log _{2} 3-\log _{2} 4=1.5850-\log _{2} 2^{2}=1.5850-2=-0.4150$

Example 12.

Write each expression as a single logarithm.

(a)  $2+3 \log _{5} x^{2}$

(b)  $\log _{3} 2+\log _{9} 81$

(c)  $\log _{b}(3 x)+\log _{b}(4 y)-\log _{b}(2 z)$

(d)  $-\log _{3}(2 s)+\dfrac{1}{2} \log _{3}\left(4 t v^{3}\right)-2 \log _{3}(5 u)$

$\begin{aligned} \text{ (a) }\quad 2+3 \log _{5} x^{2}&=\log _{5} 5^{2}+\log _{5}\left(x^{2}\right)^{3}\\\\ &=\log _{5} 25+\log _{5} x^{6}\\\\ &=\log _{5} 25 x^{6} \end{aligned}$ $\begin{aligned} \text{ (b) }\quad \log _{3} 2+\log _{9} 81&=\log _{3} 2+\log _{9} 9^{2}\\\\ &=\log _{3} 2+2 \end{aligned}$ $\begin{aligned} \text{ (c) }\quad \log _{b}(3 x)+\log _{b}(4 y)-\log _{b}(2 z)&=\log _{b}\left(\dfrac{3 x \times 4 y}{2 z}\right)\\\\ &=\log \left(\dfrac{6 x y}{z}\right) \end{aligned}$ $\begin{aligned} \text{ (d) }\quad-\log _{3}(25)+\dfrac{1}{2} \log _{3}\left(4 t v^{4}\right)-2 \log _{3}(5 u)&=-\log _{3}(25)+\log _{3}\left(4 t v^{4}\right)^{\frac{1}{2}}-\log _{3}(5 u)^{2}\\\\ &=\log _{3}\left(\dfrac{\left(4 t v^{4}\right)^{\frac{1}{2}}}{25 x(5 u)^{2}}\right)\\\\ &=\log _{3} \dfrac{t^{\frac{1}{2}} v^{2}}{255 u^{2}} \end{aligned}$

Example 13.

Solve the following equations for $x$.

$\log _{2}\left(3 x^{2}-1\right)-\log _{2}(2 x)=0$

$\begin{aligned} \log _{2}\left(3 x^{2}-1\right)&=\log _{2}(2 x) \\\\ 3 x^{2}-1&=2 x \\\\ 3 x^{2}-2 x-1&=0 \\\\ (3 x+1)(x-1)&=0 \end{aligned}$ $3 x+1=0$  (or)   $x-1=0 $ $x=-\dfrac{1}{3}$  (or)   $x=1$ But $x=-\dfrac{1}{3}$ is impossible because $\log _{2}\left(-\dfrac{1}{3}\right)$ is not defined. So $x=1$ is the only solution.

Example 14.

Suppose that $\log _{b}\left(x y^{2}\right)=4$ and $\log _{b}\left(\dfrac{x^{3}}{y}\right)=5$.

(a) Write the equation connecting $\log _{b} x$ and $\log _{b} y$.

(b) Find the values of $\log _{b} x$ and $\log _{b} y$.

(c) Find $\log _{b}\left(y^{5} \sqrt{x}\right)$.

(d) Write $x$ and $y$ in terms of $b$.

$\begin{aligned} \text{ (a) Since }\quad\log _{b}\left(x y^{2}\right)&=4\\\\ \log _{b} x+\log _{b} y^{2}&=4 \\\\ \log _{b} x+2 \log _{b} y&=4-----(1) \end{aligned}$ $\begin{aligned} \text{ (b) Since }\quad \log _{b}\left(\dfrac{x^{3}}{y}\right)&=5\\\\ \log _{b} x^{3}-\log _{b} y&=5 \\\\ 3 \log _{b} x-\log _{b} y&=5 \\\\ 6 \log _{b} x-2 \log _{b} y&=10-----(2) \end{aligned}$ $e q^{n:}(1)+(2)$, we get $\begin{aligned} 7 \log _{b} x &=14 \\\\ \log _{b} x &=2 \end{aligned}$ Substituting $\log _{b} x=2$ in $e q^{n:}(2)$ $\begin{aligned} 2+2 \log _{b} y&=4\\\\ 2 \log _{b} y&=2 \\\\ \log _{b} y&=1 \end{aligned}$ $\begin{aligned} \text { (c) } \quad\log _{b}\left(y^{5} \sqrt{x}\right) &=\log _{b} y^{5}+\log _{b} \sqrt{x} \\\\ &=5 \log _{b} y+\log _{b} x^{\frac{1}{2}} \\\\ &=5(1)+\dfrac{1}{2} \log _{b} x \\\\ &=5+\dfrac{1}{2}(2) \\\\ &=6 \end{aligned}$ (d) Since $\log _{b} y=1$, we have $y=b$    Since $\log _{b} x=2$, we have $x=b^{2}$

3.4 Change of Base

Logarithms to any base can be computed by using logarithms to some other bases. $N$, $a$ နှင့် $b$ တို့သည်အပေါင်းကိန်းများဖြစ်ကာ $1$ မဟုတ်ပါက logarithm တစ်ခု၏အခြေနှစ်မျိုးကိုစိတ်ကြိုက်ပြောင်းပေးနိုင်သည်။ Examples $\log _{a} N=\dfrac{\log _{b} N}{\log _{b} a}$,  $\log _{a} N =\dfrac{\log _{N} N}{\log _{N} a}$,  $a^{\log_{k} b}=b^{\log_{k} a}$ Theorem 2 Suppose $a$ and $b$ are any two positive real numbers other than $1$. If $N$ is any positive real number, then $L 8. \log _{a} N=\dfrac{\log _{b} N}{\log _{b} a}$ Proof: Since $N=a^{\log _{a} N}$, we have $\begin{gathered} \log _{b} N=\log _{b} a^{\log _{a} N} \\\\ \log _{b} N=(\log _{a} N)(\log _{b} a) \\\\ \log _{a} N=\dfrac{\log _{b} N}{\log _{b} a} \end{gathered}$ Corollary 1 If $a$ and $N$ are any two positive real numbers other than $1$, then $L 9.  \log _{a} N=\dfrac{1}{\log _{N} a}$ Proof: Using $L 8$ and $L 3$, $\begin{aligned} \log _{a} N &=\dfrac{\log _{N} N}{\log _{N} a} \\\\ &=\dfrac{1}{\log _{N} a} \end{aligned}$ Corollary 2 If $ a,p,N $ are any positive real numbers such that $a\neq1$ aand $a^{p}\neq1$, then $L 10.  \log _{a^{p}} N=\dfrac{1}{p} \log _{a} N$ Proof: Using $L9$ and $L 6$, $\begin{aligned} \log _{a^{p}} N&=\dfrac{\log _{N} N}{\log _{N} a^{p}}\\\\ &=\dfrac{1}{p \log _{N} a} \\\\ &=\dfrac{1}{p} \log _{a} N \end{aligned}$ Corollary 3 If $ a,b,k $ are any positive real numbers other than $1$, then $L 11.  a^{\log _{k} b}=b^{\log _{k} a}$ Proof: Using $L 2$ and $L 6$, $\begin{aligned} a^{\log_{k} b} &=a^{\log_{k}(a^{\log_{a} b})}\\\\ &=a^{(\log_{a} b)(\log_{k} a)}\\\\ &=(a^{\log_{a} b})^{\log_{k} a}\\\\ &=b^{\log_{k} a} \end{aligned}$

Example 15.

Given that $\log _{p} x=20$ and $\log _{p} y=5$, find $\log _{y} x$ and $\log _{x} y$.

$\begin{aligned} \log _{y} x&=\dfrac{\log _{p} x}{\log _{p} y}=\dfrac{20}{5}=4 \\\\ \log _{x} y&=\dfrac{\log _{p} y}{\log _{p} x}=\dfrac{5}{20}=\dfrac{1}{4} \end{aligned}$

Example 16.

Find the value of (a)  $2^{\frac{\log _{5} 3}{\log _{5} 2}}$

(b)  $5^{\frac{1}{\log _{7} 5}}$

(c)  $\log _{3} 5 \times \log _{25} 27$

(a) $2^{\frac{\log _{5} 3}{\log _{5} 2}}=2^{\log _{2} 3}=3$ (b)  $5^{\frac{1}{\log _{7} 5}}=5^{\log _{5} 7}=7$ $\begin{aligned} \text{ (c) }\quad \log _{3} 5 \times \log _{25} 27&=\log _{3} 5 \times \log_{5^{2}} 3^{3}\\\\ &=\log _{3} 5 \times \dfrac{3}{2} \times \log _{5} 3 \\\\ &=\log _{3} 5 \times \dfrac{3}{2} \times \dfrac{1}{\log _{3} 5} \\\\ &=\dfrac{3}{2} \end{aligned}$

Example 17.

Solve the equation $\log _{3} x=3-2 \log _{x^{3}}$, where $x>0$ and $x \neq 1$.

$\begin{aligned} \log _{3} x&=3-2 \log _{x} 3\\\\ \log _{3} x&=3-\dfrac{2}{\log _{3} x}\\\\ \left(\log _{3} x\right)^{2}&=3 \log _{3} x-2\\\\ \left(\log _{3} x\right)^{2}-3 \log _{3} x+2&=0\\\\ \left(\log _{3} x-2\right)\left(\log _{3} x \quad 1\right)&=0\\\\ \log _{3} x-2=0 \text{ (or) } \log _{3} x-1&=0\\\\ \log _{3} x =2 \text { (or) } \log _{3} x&=1 \\\\ x =3^{2} \text { (or) } x&=3 \\\\ x =9 \text { (or) } x&=3 \end{aligned}$

Example 18.

Solve the logarithmic equation $\log _{3} x=\log _{9}(x+6)$.

$\begin{aligned} \log _{3} x&=\log _{9}(x+6) \\\\ \log _{3} x&=\dfrac{\log _{3}(x+6)}{\log _{3} 9} \\\\ \log _{3} x \cdot \log _{3} 9&=\log _{3}(x+6) \\\\ \log _{3} x \cdot \log _{3} 3^{2}&=\log _{3}(x+6) \\\\ 2 \log _{3} x&=\log _{3}(x+6) \\\\ \log _{3} x^{2}&=\log _{3}(x+6) \\\\ x^{2}&=x+6 \\\\ x^{2}-x-6&=0\\\\ (x-3)(x+2)&=0 \\\\ x-3=0(\text { or }) x+2&=0 \\\\ x=3 \text { (or) } x&=-2 \end{aligned}$ But $x=-2$ is impossible. $\therefore x=3$.

3.5 Common Logarithm and Natural Logarithm

Common Logarithm The logarithm of $N$ to the base $10$ is said to be a common logarithm, and is usually written as $\log N$. အခြေ $10$ အသုံးပြုထားသော logarithm ကို common logarithm ဟုခေါ်သည်။ $\log N$ ဟုရေးသားဖော်ပြသည်။အခြေ $10$ ကိုချန်လှပ်၍ရေးလေ့ရှိသည်။ $\begin{aligned} \text{ If }\quad N&=a\times 10^{n},\\\\ \text{ then } \log N&=\log\left(a\times10^{n}\right)\\\\ &=\log 10^{n}+\log a\\\\ &=n + \log a \end{aligned}$ In this case $n$ and $\log a$ respectively said to be the characteristics and mantissa of $\log N$. အပေါင်းကိန်းစစ်တစ်ခု $N$ ၏ logarithm တန်ဖိုးရှာလိုသည့်အခါပထမဦးစွာ $N$ ကို scientific notation သို့ပြောင်းရမည်။ ထို့နောက်ညီမျှခြင်းနှစ်ဖက်စလုံးကို logarithm ယူခြင်းဖြင့် $n + \log a$ ပုံစံသို့ရောက်ရှိမည်ဖြစ်သည်။ $n$ ကို $\log N$ ၏ characteristics ဟုခေါ်ပြီး $\log a$ ကို $\log N$ ၏ mantissa ဟုခေါ်သည်။ Euler's Number As a positive integer $n$ become very large, the value of $\left(1+\dfrac{1}{n}\right)^{n}$ approaches an irrational number, which is denoted by $e$. The number $e$ is called Euler's number. အပေါင်းကိန်းတစ်ခုဖြစ်သည့် $n$ ၏တန်ဖိုးသည်အလွန်ကြီးလာသည့်အခါ $\left(1+\dfrac{1}{n}\right)^{n}$ ၏တန်ဖိုးသည်အီရာရှင်နယ်ကိန်းတစ်ခုနားသို့ချဉ်းကပ်သွား၏။ ၎င်းအီရာရှင်နယ်ကိန်းကို Euler's number ဟုခေါ်၍ $e$ ဖြင့်မှတ်သားဖော်ပြသည်။ $e\approx2.71828$ Natural Logarithm The logarithm of $N$ to the base $e$ is called a natural logarithm and is denoted by $\ln N$. အခြေ $e$ အသုံးပြုထားသော logarithm ကို natural logarithm ဟုခေါ်သည်။ $\ln N$ ဟုရေးသားဖော်ပြသည်။ $\log_{e} N=\ln N$

Example 19.

Given that $\log _{10} 7=0.8451$; what are the characteristics and the mantissas of $\log _{10} 0.007$ and $\log _{10} 700$ ?

$\begin{aligned} \log _{10} 7&=0.8451 \\\\ \log _{10} 0.007 &=\log _{10}\left(7 \times 10^{-3}\right) \\\\ &=\log _{10} 7+\log _{10} 10^{-3} \\\\ &=0.8451-3 \\\\ &=-3+0.8451 \end{aligned}$ The characteristics is $-3$ and the mantissa is $0.8451$. $\begin{aligned} \log _{10} 700 &=\log _{10}\left(7 \times 10^{2}\right) \\\\ &=\log _{10} 7+\log _{10} 10^{2} \\\\ &=0.8451+2 \\\\ &=2+0.8451 \end{aligned}$ The charasteristics is 2 and the mantissa is $0.8451$.

Example 20.

Two nonnegative real numbers $A$ and $P$ are related by the formula $A=Pe^{0.085t}$. Given that $\ln 2=0.6931$, find the value of $t$ for which $A$ becomes$200%$ of $P$.

$\begin{aligned} A &=P e^{0.085 t} \\\\ A&=200 \% \text { of } P \\\\ 200 \% P &=P e^{0.085 t} \\\\ \dfrac{200}{100} \times P&=P e^{0.085 t} \\\\ 2 &=e^{0.085 t} \\\\ \log_{e}2&=0.085 t \\\\ \ln 2&=0.085 t \\\\ t &=\dfrac{\ln 2}{0.085} \\\\ &= \dfrac{0.6931}{0.0850} \\\\ &=\dfrac{6931}{850} \end{aligned}$

Example 21.

Given that $\log _{10} 2=0.3010$, $\log _{10}9.87=0.9943$ and $\log _{10}8.5=0.9294$; evaluate $\dfrac{200 \times98.7 \times 85}{8.5^{3}}$ by using logarithms.

$\begin{aligned} Let p&=\dfrac{200 \times 98.7 \times 85}{8.5^{3}}\\\\ \log _{10} p&=\log _{10} \dfrac{200 \times 98.7 \times 85}{8.5^{3}}\\\\ &=\log _{10} 200+\log _{10} 98.7+\log _{10} 85-\log _{10} 8.5^{3}\\\\ &=\log _{10}\left(2 \times 10^{2}\right)+\log _{10}\left(9.87 \times 10^{-1}\right)+\log _{10}\left(8.5 \times 10^{-1}\right)-3 \log _{10} 8.5\\\\ &=\log _{10} 2+2+\log _{10} 9.87+\log _{10} 10^{-1}+\log _{10} 8.5+\log _{10^{10}} 10^{-1}-3(0.9294)\\\\ &=0.3010+2+0.9943-1+0.9294-1-2.7882\\\\ &=5.2953-1.8588\\\\ &=3.4365\\\\ p&=10^{3.4365} \end{aligned}$ $\therefore \dfrac{200 \times 98.7 \times 85}{8.5^{3}}=10^{3.4365}$

Exercise 3.1

1. How many significant figures are there in each of the following numbers?

(a) $2.175$  (b) $0.2175$  (c) $0.0075$

(d) $89400$  (e) $0.00046$





(a)  $2.175$  (four significant figures) (b)  $0.2175$  (four significant figures) (c)  $0.0075$ (two significant figures) (d)  $89400$  (five significant figures) (e)  $0.00046$  (two significant figures)




2. Write in scientific notation.

(a) $24.86$  (b) $2.486$  

(c) $0.2486$  (d) $0.002486$

(e) $0.073$  (f) $0.0086$

(g) $0.934$  (h) $7$

(i) $0.00056857$  (j) $6.843250$





(a)  $24.86=2.486 \times 10^{1}$ (b)  $2.486=2.486 \times 10^{0}$ (c)  $0.2486=2.486 \times 10^{-1}$ (d)  $0.002486=2.486 \times 10^{-3}$ (e)  $0.073=7.3 \times 10^{-2}$ (f)  $0.0086=8.6 \times 10^{-3}$ (g)  $0.934=9.34 \times 10^{-1}$ (h)  $7=7.0 \times 10^{0}$ (i)  $0.00056857=5.6857 \times 10^{-4}$ (j)  $6.843250=6.843250 \times 10^{0}$



3. Write each number in ordinary decimal form.

(a)  $7.84\times 10^{4}$   (b)  $7.89\times 10^{-4}$

(c)  $2.25\times 10^{5}$   (d)  $4.01\times 10^{-3}$





(a)  $7.84 \times 10^{4}=78400$ (b)  $7.89 \times 10^{-4}=0.000789$ (c)  $2.25 \times 10^{5}=225000$ (d)  $4.01 \times 10^{-3}=0.00401$



4. Simplify and give the answers in scientific notation.

(a)  $2.3 \times 10^{2}+1.7 \times 10^{2}$

(b)  $4.6 \times 10^{-3}-2.5 \times 10^{-3}$

(c)  $\left(4.5 \times 10^{6}\right) \times\left(1.5 \times 10^{-2}\right)$

(d)  $\dfrac{7.6 \times 10^{5}}{1.9 \times 10^{-2}}$





(a)  $2.3 \times 10^{2}+1.7 \times 10^{2}=(2.3+1.7) \times 10^{2}$ $=4.0 \times 10^{2}$ (b)  $4.6 \times 10^{-3}-2.5 \times 10^{-3}=(4.6-2.5) \times 10^{-3}=2.1 \times 10^{-3}$ (c)  $\left(4.5 \times 10^{6}\right) \times\left(1.5 \times 10^{-2}\right)=(4.5 \times 1.5) \times 10^{6-2}=6.75\times 10^{4}$ (d)  $\dfrac{7.6 \times 10^{5}}{1.9 \times 10^{-2}}=\dfrac{76}{19} \times 10^{5+2}=4.0 \times 10^{7}$



5. Compute using scientific notation.

(a)  $\dfrac{2.5 \times 10^{2}}{0.25 \times 0.002}$

(b)  $\dfrac{33,000,000 \times 0.4}{1.1 \times 30}$

(c)  $\dfrac{50 \times 0.014 \times 0.30}{10500}$

(d)  $\dfrac{7000 \times 80 \times 300}{400}$





$\begin{aligned} \text { (a) }\quad \dfrac{2.5 \times 10^{2}}{0.25 \times 0.002}&=\dfrac{2.5 \times 10^{2}}{\left(2.5 \times 10^{-1}\right) \times\left(2 \times 10^{-3}\right)}\\\\ &=\dfrac{10^{2}}{2 \times 10^{-4}}\\\\ &=\dfrac{1}{2} \times 10^{6}\\\\ &=0.5 \times 10^{6}\\\\ &=5 \times 10^{5}\\\\ &=500000\\\\ \end{aligned}$ $\begin{aligned} \text { (b) }\quad \dfrac{33,000,000 \times 0.4}{1.1 \times 30}&=\dfrac{\left(3.3 \times 10^{7}\right) \times\left(4 \times 10^{-1}\right)}{\left(1.1 \times 10^{0}\right) \times\left(3 \times 10^{1}\right)}\\\\ &=4 \times 10^{5}\\\\ &=400000 \end{aligned}$ $\begin{aligned} \text { (c) }\quad \dfrac{50 \times 0.014 \times 0.30}{10500}&=\dfrac{\left(5.0 \times 10^{1}\right) \times\left(1.4 \times 10^{-2}\right) \times\left(3.0 \times 10^{-1}\right)}{1.0500 \times 10^{4}}\\\\ &=\dfrac{21 \times 10^{-2}}{1.05 \times 10^{4}}\\\\ &=20 \times 10^{-6}\\\\ &=0.00002 \end{aligned}$ $\begin{aligned} \text { (d) }\quad \dfrac{7000 \times 80 \times 300}{400} &=\dfrac{\left(7 \times 10^{3}\right) \times\left(8 \times 10^{1}\right) \times\left(3 \times 10^{2}\right)}{4 \times 10^{2}} \\\\ &=42 \times 10^{4} \\\\ &=420000 \end{aligned}$

Exercise 3.2

1. Write the following equations in Logarithmic form.

(a) $ 3^{4} =81$     (b) $ 9^{\frac{3}{2}}=27$

(c) $10^{-3}=0.001$  (d) $3^{-1}=\dfrac{1}{3}$

(e) $\left(\dfrac{1}{4}\right)^{-3} =64$





$\begin{aligned} \text{ (a) }\quad \quad 3^{4} &=81 \\\\ \log _{3} 81&=4 \end{aligned}$ $\begin{aligned} \text{ (b) }\quad\quad 9^{\frac{3}{2}}&=27\\\\ \log _{9} 27&=\dfrac{3}{2} \end{aligned}$ $\begin{aligned} \text{ (c) }\quad\quad 10^{-3}&=0.001\\\\ \log _{10} 0.001&=-3 \end{aligned}$ $\begin{aligned} \text{ (d) }\quad\quad 3^{-1}&=\dfrac{1}{3}\\\\ \log _{3} \dfrac{1}{3}&=-1 \end{aligned}$ $\begin{aligned} \text { (e) }\quad\quad \left(\dfrac{1}{4}\right)^{-3} &=64 \\\\ \log _{\frac{1}{4}} 64 &=-3 \end{aligned}$



2. Write the following equations in exponential form.

(a) $ \log _{10} 3=0.4771$

(b) $\log _{6} 0.001=-3.855$

(c) $ \log _{44} 12=\dfrac{1}{2}$

(d) $-5=\log _{3} \dfrac{1}{243}$

(e) $\log _{x} 7=y^{2}$, where, $0<x<1$





$\begin{aligned} \text{ (a) } \log _{10} 3&=0.4771\\\\ 10^{0.4771}&=3 \end{aligned}$ $\begin{aligned} \text{ (b) } \log _{6} 0.001&=-3.855\\\\ 6^{-3.855}&=0.001 \end{aligned}$ $\begin{aligned} \text{ (c) } \log _{44} 12&=\dfrac{1}{2}\\\\ 144^{\frac{1}{2}}&=12 \end{aligned}$ $\begin{aligned} \text{ (d) } -5&=\log _{3} \dfrac{1}{243}\\\\ 3^{-5}&=\dfrac{1}{243} \end{aligned}$ $\begin{aligned} \text{ (e) } \log _{x} 7&=y^{2}, \text{ where }, 0<x<1 \\\\ x^{y^{2}}&=7 \end{aligned}$



3. Solve the following equations.

(a) $ \log _{7} 49=x$  (b) $\log _{x} 10=1$

(c) $ \log _{\sqrt{3}} x=2$  (d) $\log _{\sqrt{3}} x=2$

(e) $x^{\log _{x} x}=5$





$\begin{aligned} \text{ (a) } \log _{7} 49&=x\\\\ \log _{7} 7^{2}&=x\\\\ 7^{x}&=7^{2}\\\\ \therefore x&=2 \end{aligned}$ $\begin{aligned} \text{ (b) } \log _{x} 10&=1\\\\ x^{1}&=10\\\\ x&=10 \end{aligned}$ $\begin{aligned} \text{ (c) } \log _{\sqrt{3}} x&=2\\\\ \sqrt{3}^{2}&=x\\\\ \therefore x&=3 \end{aligned}$ $\begin{aligned} \text{ (d) } x^{\log _{x} x}&=5\\\\ \therefore x&=5 \end{aligned}$



4. Evaluate.

(a) $9^{\log _{9} 2}+3^{\log _{3} 8}$

(b) $\log _{4} 4^{5} \times \log _{10} 10^{2}$

(c) $7^{\log _{7} 9}+\log _{2}\left(\dfrac{1}{2}\right)$

(d) $\log _{\frac{1}{2}}\left(\dfrac{1}{8}\right)-4 \log _{10} 10$

(e) $10^{1-\log _{10} 3}$





(a)  $9^{\log _{9} 2}+3^{\log _{3} 8}=2+8=10$ (b)  $\log _{4} 4^{5} \times \log _{10} 10^{2}=5 \times 2=10$ (c)  $7^{\log _{7} 9}+\log _{2}\left(\dfrac{1}{2}\right)=9+\log _{2} 2^{-1}=9-1=8$ (d)  $\log _{\frac{1}{2}}\left(\dfrac{1}{8}\right)-4 \log _{10} 10=\log _{\frac{1}{2}}\left(\dfrac{1}{2}\right)^{3}-4=3-4=-1$ (e)  $10^{1-\log _{10} 3}=\dfrac{10}{10^{\log _{10} 3}}=\dfrac{10}{3}=3 \dfrac{1}{3}$



5. PFind the value of $x$ in each of the following problems.

(a) $\log _{3}(2 x-5)=2$,   where   $x>\dfrac{5}{2}$

(b) $\log _{77}\left(\log _{7} x\right)=0$,  where  $x>0$

(c) $8+3^{x}=10$, given that $\log_{3}4=1.2619$





$\begin{aligned} \text{ (a) } \log _{3}(2 x-5)&=2, \text{ where } x>\dfrac{5}{2}\\\\ 2 x-5 &=3^{2} \\\\ 2 x-5 &=9 \\\\ 2 x &=14 \\\\ x &=7 \end{aligned}$ $\begin{aligned} \text{ (b) } \log _{77}\left(\log _{7} x\right)&=0, \text{ where } x>0\\\\ \log _{7} x &=77^{0} \\\\ \log _{7} x &=1 \\\\ x &=7^{1} \end{aligned}$ $\begin{aligned} \text{ (c) }\log _{3} 2 &=0.6309 \\\\ 8+3^{x} &=10 \\\\ 3^{x} &=2 \\\\ \log _{3} 2 &=x \\\\ \therefore x &=0.6309 \end{aligned}$

Exercise 3.3

1. Replace $□$ with the appropriate number.

(a) $\log _{3} 24=\log _{3} 6+\log _{3} □$

(b) $\log _{5} 24=\log _{5} 60+\log _{5} □$

(c) $\log _{2} □=3 \log _{2} 3$

(d) $\log _{10} 9=□ \log _{10} 3$

(e) $\log _{8} 5=\log _{8} □-\log _{8} 11$





(a)  $\log _{3} 24=\log _{3} 6+\log _{3}4$ (b)  $\log _{5} 24=\log _{5} 60+\log _{5} \dfrac{2}{5}$ (c)  $\log _{2} 27=3 \log _{2} 3$ (d)  $\log _{10} 9=2 \log _{10} 3$ (e)  $\log _{8} 5=\log _{8} 55-\log _{8} 11$



2. Write each expression as a single logarithm.

(a) $\log _{b} 20+\log _{b} 57-\log _{b} 114$

(b) $3 \log _{b} 8-\log _{b} 12=\log _{b} 8^{3}-\log _{b} 12$

(c) $\log _{b} x-2 \log _{b} y-\log _{b} a$

(d) $\log _{2} 3+\log _{4} 15$





(a)  $\log _{b} 20+\log _{b} 57-\log _{b} 114=\log _{b} \dfrac{20 \times 57}{114}=\log _{b} 10$ (b)  $3 \log _{b} 8-\log _{b} 12=\log _{b} 8^{3}-\log _{b} 12=\log _{b} \dfrac{8^{3}}{12}=\log _{b} \dfrac{128}{3}$ (c)  $\log _{b} x-2 \log _{b} y-\log _{b} a=\log _{b} x-\log _{b} y^{2}-\log _{b} a=\log _{b} \dfrac{x}{a y^{2}}$ $\begin{aligned} \text{ (d) }\log _{2} 3+\log _{4} 15&=\log _{4 ^{\frac{1}{2}}} 3+\log _{4} 15\\\\ &=\dfrac{1}{\dfrac{1}{2}} \log _{4} 3+\log _{4} 15 \\\\ &=2 \log _{4} 3+\log _{4} 15 \\\\ &=\log _{4} 3^{2}+\log _{4} 15 \\\\ &=\log _{4}\left(3^{2} \times 15\right) \\\\ &=\log _{4} 135 \end{aligned}$



3. Write each expression in terms of $\log_{2}$, $\log_{3}$, $\log_{5}$.

(a) $ \log _{b} 8$   (b) $\log _{b} 15$

(c) $\log _{b} 270$  (d) $\log _{b} \dfrac{27 \sqrt[3]{5}}{16}$

(e) $\log _{b} \dfrac{216}{\sqrt[3]{32}}$  (f) $\log _{b}(648 \sqrt{125})$





(a) $\log _{b} 8=\log _{b} 2^{3}=3 \log _{b} 2$ (b) $\log _{b} 15=\log _{b}(3 \times 5)=\log _{b} 3+\log _{b} 5$ $\begin{aligned} \text{ (c) }\quad \log _{b} 270&=\log _{b}\left(2 \times 3^{2} \times 5\right)\\\\ &=\log _{b} 2+\log _{b} 3^{3}+\log _{b} 5\\\\ &=\log _{b} 2+3 \log _{b} 3+\log 5 \end{aligned}$ $\begin{aligned} \text{ (d) }\quad \log _{b} \dfrac{27 \sqrt[3]{5}}{16}&=\log _{b}\left(\dfrac{3^{3} \times 5^{\frac{1}{3}}}{2^{4}}\right)\\\\ &=\log _{b} 3^{3}+\log _{b} 5^{\frac{1}{3}}-\log _{b} 2^{4} \\\\ &=3 \log _{b} 3+\dfrac{1}{3} \log _{b} 5-4 \log _{b} 2 \end{aligned}$ $\begin{aligned} \text{ (e) }\quad \log _{b} \dfrac{216}{\sqrt[3]{32}} &=\log _{b} \dfrac{6^{3}}{\left(2^{5}\right)^{\frac{1}{3}}} \\\\ &=\log _{b} \dfrac{(2 \times 3)^{3}}{2^{\frac{5}{3}}} \\\\ &=\log _{b}\left(\frac{2^{3} \times 3^{3}}{2^{\frac{5}{3}}}\right) \\\\ &=\log _{b}\left(2^{3-\frac{5}{3}} \times 3\right) \\\\ &=\log _{b} 2^{\frac{4}{3}}+\log _{b} 3^{3} \\\\ &=\dfrac{4}{3} \log _{b} 2+3 \log _{b} 3 \end{aligned}$ $\begin{aligned} \text { (f) } \quad\log _{b}(648 \sqrt{125}) &=\log _{b}\left(2^{3} \times 3^{4} \times\left(5^{3}\right)^{\frac{1}{2}}\right) \\\\ &=\log _{b} 2^{3}+\log _{b} 3^{4}+\log _{b} 5^{\frac{3}{2}} \\\\ &=3 \log _{b} 2+4 \log _{b} 3+\dfrac{3}{2} \log _{b} 5 \end{aligned}$



4. Evaluate each expression.

(a) $ \log _{7} 49=x$  (b) $\log _{x} 10=1$

(c) $ \log _{\sqrt{3}} x=2$  (d) $\log _{\sqrt{3}} x=2$

(e) $x^{\log _{x} x}=5$   (f) $\log _{\sqrt{3}} x=2$

(g) $x^{\log _{x} x}=5$   (h) $\log _{\sqrt{3}} x=2$





(a) $\log _{2} 128=\log _{2} 2^{7}=7$ (b) $\log _{3} 81^{4}=\log _{3}\left(3^{4}\right)^{4}=\log _{3} 3^{16}=16$ (c) $\log _{\frac{1}{2}} 8=\log _{\frac{1}{2}} 2^{3}=\log _{\frac{1}{2}}\left(\dfrac{1}{2}\right)^{-3}=-3$ (d) $\log _{8} 2=\log _{8}\left(2^{3}\right)^{\frac{1}{3}}=\log _{8} 8^{\frac{1}{3}}=\dfrac{1}{3}$ (e) $\log _{3} \dfrac{\sqrt{3}}{81}=\log _{3} \dfrac{3^{\frac{1}{2}}}{3^{4}}=\log _{3} 3^{-\frac{7}{2}}=-\dfrac{7}{2}$ (f) $\dfrac{\log _{3} \sqrt{3}}{\log _{3} 81}=\dfrac{\log _{3} 3^{\frac{1}{2}}}{\log _{3} 3^{4}}=\dfrac{\dfrac{1}{2}}{4}=\dfrac{1}{8}$ (g) $\dfrac{\log _{2} 25}{\log _{2} 5}=\dfrac{\log _{2} 5^{2}}{\log _{2} 5}=\dfrac{2 \log _{2} 5}{\log _{2} 5}=2$ (h) $\log _{4} 8=\log _{4} 2^{3}=\log _{4}\left(2^{2}\right)^{\frac{3}{2}}=\log _{4} 4^{\frac{3}{2}}=\frac{3}{2}$



5. Use $\log_{2}=0.3010$ and $\log_{3}=0.4771$ to evaluate each of the following expressions.

(a) $\log _{10} 6$   (b) $\log _{10} 1.5$

(c) $ \log _{10} \sqrt{3}$  (d) $\log _{10} 4$

(e) $\log _{10} 4.5$  (f) $\log _{10} 8$

(g) $\log _{10} 18$  (h) $\log _{10} 5$





$\log _{10} 2=0.3010, \log _{10} 3=0.4771$ $\begin{aligned} \text{ (a) }\quad \log _{10} 6&=\log _{10}(2 \times 3)\\\\ &=\log_{10}2+\log_{10}3\\\\ &=0.3010+0.4771\\\\ &=0.7781 \end{aligned}$ $\begin{aligned} \text{ (b) }\quad \log _{10} 1.5&=\log _{10}\left(\frac{3}{2}\right)\\\\ &=\log_{10}3+\log_{10}2\\\\ &=0.4771-0.3010\\\\ &=0.1761 \end{aligned}$ $\begin{aligned} \text{ (c) } \quad\log _{10} \sqrt{3}&=\log _{10} 3^{\frac{1}{2}}\\\\ &=\dfrac{1}{2}\log _{10} 3\\\\ &=\dfrac{1}{2}(0.4771)\\\\ &=0.23855 \end{aligned}$ $\begin{aligned} \text{ (d) }\quad \log _{10} 4&=\log _{10} 2^{2}\\\\ &=2\log _{10} 2\\\\ &=2(0.3010)\\\\ &=0.6020 \end{aligned}$ $\begin{aligned} \text{ (e) } \quad\log _{10} 4.5&=\log _{10}\left(\frac{9}{2}\right)\\\\ &=\log _{10} \dfrac{3^{2}}{2} \\\\ &=\log _{10} 3^{2}-\log_{10}2 \\\\ &=2 \log _{10} 3-\log_{10}2 \\\\ &=2(0.4771)-0.3010 \\\\ &=0.6532 \end{aligned}$ $\begin{aligned} \text { (f) } \quad\log _{10} 8 &=\log _{10} 2^{3} \\\\ &=3 \log _{10} 2 \\\\ &=3(0.3010) \\\\ &=0.9030 \end{aligned}$ $\begin{aligned} \text { (g) }\quad \log _{10} 18 &=\log _{10}\left(2 \times 3^{2}\right) \\\\ &=\log _{10} 2+\log _{10} 3^{2} \\\\ &=\log _{10} 2+2 \log _{10} 3 \\\\ &=0.3010+2(0.4771) \\\\ &=0.3010+0.9542\\\\ &=1.2552 \end{aligned}$ $\begin{aligned} \text { (h) }\quad \log _{10} 5 &=\log _{10}\left(\frac{10}{2}\right) \\\\ &=\log _{10} 10-\log _{10} 2 \\\\ &=1-0.3010 \\\\ &=0.6990 \end{aligned}$



6. Solve the following equations for $x$.

(a)   $\log _{a} \dfrac{18}{5}+\log _{a} \dfrac{10}{3}-\log _{a} \dfrac{6}{7}=\log _{a} x$

(b)   $\log _{b} x=2-a+\log _{b}\left(\dfrac{a^{2} b^{a}}{b^{2}}\right)$

(c)   $\log x^{3}-\log x^{2} =\log 5 x-\log 4 x$

(d)   $\log _{10} x+\log _{10} 3 =\log _{10} 6$

(e)   $8 \log x=\log a^{\frac{3}{2}}+\log 2-\dfrac{1}{2} \log a^{3}-\log \dfrac{2}{a^{4}}$





$\begin{aligned} \text { (a) }\quad \log _{a} \dfrac{18}{5}+\log _{a} \dfrac{10}{3}-\log _{a} \dfrac{6}{7}&=\log _{a} x\\\\ \log _{a}\left(\dfrac{18}{5} \times \dfrac{10}{3}\right)-\log _{a} \dfrac{6}{7}&=\log _{a} x\\\\ \log _{a} 12-\log _{a}\left(\dfrac{7}{6}\right)^{-1}&=\log _{a} x\\\\ \log _{a} 12+\log _{a} \dfrac{7}{6}&=\log _{a} x\\\\ \log _{a}\left(12 \times \dfrac{7}{6}\right)&=\log _{a} x\\\\ \log _{a} 14&=\log _{a} x\\\\ \therefore x&=14 \end{aligned}$ $\begin{aligned} \text { (b) }\quad \log _{b} x&=2-a+\log _{b}\left(\dfrac{a^{2} b^{a}}{b^{2}}\right)\\\\ \log _{b} x&=2-a+\log _{b} a^{2}+\log _{b} b^{a}-\log _{b} b^{2}\\\\ \log _{b} x&=2-a+\log _{b} a^{2}+a-2\\\\ \log _{b} x&=\log _{b} a^{2}\\\\ \therefore x&=a^{2} \end{aligned}$ $\begin{aligned} \text { (c) }\quad \log x^{3}-\log x^{2} &=\log 5 x-\log 4 x \\\\ \log \left(\dfrac{x^{3}}{x^{2}}\right) &=\left(\log \dfrac{5 x}{4 x}\right) \\\\ \log x &=\log \dfrac{5}{4} \\\\ x &=\dfrac{5}{4} \end{aligned}$ $\begin{aligned} \text { (d) }\quad \log _{10} x+\log _{10} 3 &=\log _{10} 6 \\\\ \log _{10}(x \times 3) &=\log _{10} 6 \\\\ 3 x &=6 \\\\ x &=2 \\ \end{aligned}$ $\begin{aligned} \text { (e) }\quad 8 \log x&=\log a^{\frac{3}{2}}+\log 2-\dfrac{1}{2} \log a^{3}-\log \dfrac{2}{a^{4}} \\\\ 8 \log x&=\dfrac{3}{2} \log a+\log 2-\dfrac{3}{2} \log a-\log 2+\log a^{4} \\\\ 8 \log x&= \log a^{4} \\\\ 8 \log x&=4 \log a \\\\ \log x &=\dfrac{1}{2} \log a \\\\ \log x &=\log a^{\frac{1}{2}} \\ x &=a^{\frac{1}{2}} \\\\ x &=\sqrt{a} \end{aligned}$



7. Given that $\log_{10}5=0.6990$ and $\log_{10}x=0.2330$. What is the value of $x$?





$\log _{10} 5=0.6990,\log _{10} x=0.2330$ $\begin{aligned} \log _{10} x&=0.2330 \\\\ \log _{10} x&=\dfrac{0.6990}{3} \\\\ \log _{10} x &=\dfrac{1}{3} \log _{10} 5 \\\\ \log _{10} x &=\log _{10} 5^{\frac{1}{3}} \\\\ x &=5^{\frac{1}{3}} \\\\ x &=\sqrt[3]{5} \end{aligned}$



8. Show that if $\log_{e}I=-\dfrac{R}{L}t+\log_{e}I_{0}$, then $I=I_{0}e^{-\frac{Rt}{L}}$.





$\begin{aligned} \log _{e} I&=-\dfrac{R}{L} t+\log _{e} I_{0}\\\\ \log _{e} I-\log _{e} I_{0}&=-\dfrac{R}{L} t\\\\ \log _{e}\left(\dfrac{I}{I_{0}}\right) &=-\dfrac{R t}{L} \\\\ \dfrac{I}{I_{0}} &=e^{-\frac{R t}{L}} \\\\ I &=I_{0} e^{-\frac{R t}{L}} \end{aligned}$



9. Show that if $\log_{b}y=\dfrac{1}{2}\log_{b}x+c$, then $y=b^{c}\sqrt{x}$.





$ \begin{aligned} \log _{b} y&=\dfrac{1}{2}\log _{b} x+c \\\\ \log _{b} y &=\log _{b} x^{\frac{1}{2}}+\log _{b} b^{c} \\\\ \log _{b} y &=\log _{b}\left(x^{\frac{1}{2}} \times b^{c}\right) \\\\ y &=x^{\frac{1}{2}} b^{c} \\\\ y &=\sqrt{x} b^{c} \end{aligned}$



10. Show that

(a) $\dfrac{1}{4} \log _{10} 8+\dfrac{1}{4} \log _{10} 2=\log _{10} 2$

(b) $4 \log _{10} 3-2 \log _{10} 3+1=\log _{10} 90$





$\begin{aligned} \text{ (a) }\quad \dfrac{1}{4} \log _{10} 8+\dfrac{1}{4} \log _{10} 2&=\log _{10}\left(2^{3}\right)^{\frac{1}{4}}+\log _{10} 2^{\frac{1}{4}}\\\\ &=\log _{10} 2^{\frac{3}{4}}+\log _{10} 2^{\frac{1}{4}} \\\\ &=\log _{10}\left(2^{\frac{3}{4}} \times 2^{\frac{1}{4}}\right) \\\\ &=\log _{10}\left(2^{\frac{3}{4}+\frac{1}{4}}\right)\\\\ &=\log _{10} 2 \end{aligned}$ $\begin{aligned} \text{ (b) }\quad 4 \log _{10} 3-2 \log _{10} 3+1&=\log _{10} 3^{4}-\log _{10} 3^{2}+\log _{10} 10\\\\ &=\log _{10}\left(\frac{3^{4} \times 10}{3^{2}}\right) \\\\ &=\log _{10}\left(9\times10\right)\\\\ &=\log _{10} 90 \end{aligned}$



11. Show that

(a) $a^{2 \log _{a} 3}+b^{3 \log _{b} 2}=17$

(b) $3 \log _{6} 1296=2 \log _{4} 4096$





$\begin{aligned} \text { (a) }\quad a^{2 \log _{a} 3}+b^{3 \log _{b} 2} &=a^{\log _{a} 3^{2}}+b^{\log _{b} 2^{3}} \\\\ &=3^{2}+2^{3} \\\\ &=9+8 \\\\ &=17 \end{aligned}$ $\begin{aligned} \text { (b) }\quad 3 \log _{6} 1296&=3 \log _{6} 6^{4}=3 \times 4=12 \\\\ 2 \log _{4} 4096&=2 \log _{4} 4^{6}=\log _{4}\left(4^{6}\right)^{2}=\log _{4} 4^{12}=12 \\\\ 3 \log _{6} 1296&=2 \log _{4} 4096 \end{aligned}$



12. Given that $\log_{10}12=1.0792$ and $\log_{10}24=1.3802$, deduce the values of $\log_{10}2$ and $\log_{10}6$.





$\log _{10} 12=1.0792, \log _{10} 24=1.3802 $ $\begin{aligned} \log _{10} 2&=\log _{10}\left(\dfrac{24}{12}\right) \\\\ &=\log _{10} 24-\log _{10} 12 \\\\ &=1.3802-1.0792 \\\\ &=0.3010 \end{aligned}$ $\begin{aligned} \log_{10}6&=\log _{10}\left(\dfrac{12}{2}\right)\\\\ &=\log _{10} 12-\log _{10} 2\\\\ &=1.0792-0.3010\\\\ &=0.7782 \end{aligned}$



13. If $\log_{x}a=5$ and $\log_{3a}3a=9$, find the values of $a$ and $x$.





$\log _{x} a=5, \quad \log _{x} 3 a=9$ $\begin{aligned} \log _{x} 3 a=\log _{x} 3+\log _{x} a &=9 \\\\ \log _{x} 3+5 &=9 \\\\ \log _{x} 3 &=4 \\\\ 3 &=x^{4} \\\\ x &=\sqrt[4]{3} \end{aligned}$ $\begin{aligned} \log _{x} a&=5\text{ ( given ) }\\\\ a&=x^{5}\\\\ &=(\sqrt[4]{3})^{5}\\\\ &=\sqrt[4]{3^{5}}\\\\ &=\sqrt[4]{3^{4}\cdot 3}\\\\ &=3\cdot\sqrt[4]{3} \end{aligned}$

14.(a) If $\log_{10}2=a$, find $\log_{10}8+\log_{10}25$ in terms of $a$.

  (b) If $a=10^{x}$ and $b=10^{y}$, express $\log_{10}\left(a^{4} b^{3}\right)$ in terms of $x$ and $y$.

(a) $\log _{10} 2=a$ $\begin{aligned} \log _{10} 8+\log _{10} 25 &=\log _{10} 2^{3}+\log _{10}\left(\frac{100}{4}\right) \\\\ &=3 \log _{10} 2+\log _{10}10^{2}-\log _{10} 2^{2} \\\\ &=3 \log _{10} 2+2-2 \log _{10} 2 \\\\ &=2+\log _{10} 2 \\\\ &=2+a \end{aligned}$ (b)   $a=10^{x}, b =10^{y}$ $\begin{aligned} \log _{10}\left(a^{4} b^{3}\right) &=\log _{10} a^{4}+\log _{10} b^{3} \\\\ &=4 \log _{10} a+3 \log _{10} b \\\\ &=4 \log _{10} 10^{x}+3 \log _{10} 10^{y} \\\\ &=4 x+3 y \end{aligned}$

15.(a) If $\log_{2}\left(4x - 4\right)=2$, find the value of $\log_{4}x$.

  (b) Prove that if $\dfrac{1}{2}\log_{3}M+3\log_{3}N=1$ then $MN^{6}=9$.

$\begin{aligned} \text { (a) }\quad \log _{2}(4 x-4) &=2 \\\\ 4 x-4 &=4 \\\\ 4 x &=8 \\\\ x &=2\\\\ x&=4^{\frac{1}{2}} \\\\ \log _{4} x&=\dfrac{1}{2} \end{aligned}$ $\begin{aligned} \text { (b) }\quad \dfrac{1}{2} \log _{3} M+3 \log _{3} N&=1 \\\\ \log _{3} M^{\frac{1}{2}}+\log _{3} N^{3}&=1 \\\\ \log _{3}\left(M^{\frac{1}{2}} N^{3}\right)&=1 \\\\ M^{\frac{1}{2}} N^{3}&=3 \\\\ \left(M^{\frac{1}{2}} N^{3}\right)^{2}&=3^{2} \\\\ M N^{6}=9 \end{aligned}$

Exercise 3.4

1. If $\log_{a}b+\log_{b}a^{2}=3$, find $b$ in terms of $a$.





$\begin{aligned} \log _{a} b+\log _{b} a^{2}&=3 \\\\ \log _{a} b+2 \log _{b} a&=3 \\\\ \log _{a} b+\dfrac{2}{\log _{a} b}&=3 \\\\ \left(\log _{a} b\right)^{2}+2&=3 \log _{a} b \\\\ \left(\log _{a} b\right)^{2}-3 \log _{a} b+2&=0 \\\\ \left(\log _{a} b-2\right)\left(\log _{a} b-1\right)&=0 \\\\ \log _{a} b-2=0 \text { (or) } \log _{a} b-1&=0 \\\\ \log _{a} b=2 \text { (or) }\log _{a} b&=1 \\\\ b=a^{2} \text { (or) } b&=a \end{aligned}$



2. Show that

(a) $\log_{4}x=2\log_{16}x$

(b) $\log_{b}x=3\log_{b^{3}}x$

(c) $\log_{2}x=(1+\log_{2}3)\log_{6}x$





(a)  $2 \log _{16} x=\log _{4^{2}} x^{2}=\dfrac{2}{2} \log _{4} x=\log _{4} x$ (b)  $3 \log _{b^{3}} x=\log _{b^{3}} x^{3}=\dfrac{3}{3} \log _{b} x=\log _{b} x$ $\begin{aligned} \text{ (c) }\quad \left(1+\log _{2} 3\right) \log _{6} x&=\left(\log _{2} 2+\log _{2} 3\right) \log _{6} x\\\\ &=\log _{2}(2 \times 3) \times \log _{6} x \\\\ &=\log _{2} 6 \times \log _{6} x \\\\ &=\dfrac{\log _{6}}{\log 2} \times \dfrac{\log x}{\log 6} \\\\ &=\dfrac{\log x}{\log 2} \\\\ &=\log _{2} x \end{aligned}$



3. If $a=\log_{b}c$, $b=\log_{c}a$ and $c=\log_{a}b$, prove that $a b c =1$.





$a=\log _{b} c, b=\log _{c} a, c=\log _{a} b$ $\begin{aligned} a b c&=\log _{b} c \times \log _{c} a \times \log _{a} b\\\\ &=\dfrac{\log c}{\log b} \times \dfrac{\log a}{\log c} \times \dfrac{\log b}{\log a}\\\\ &=1 \end{aligned}$



4. Show that

(a) $\left(\log _{10} 4-\log _{10} 2\right) \log _{2} 10=1$

(b) $2 \log _{2} 3\left(\log _{9} 2+\log _{9} 4\right)=3$





$\begin{aligned} \text { (a) }\quad\left(\log _{10} 4-\log _{10} 2\right) \log _{2} 10&=\log _{10}\left(\dfrac{4}{2}\right) \times \log _{2} 10\\\\ &=\log _{10} 2 \times \log _{2} 10\\\\ &=\dfrac{\log 2}{\log 10} \times \dfrac{\log 10}{\log 2}\\\\ &=1 \end{aligned}$ $\begin{aligned} \text { (b) }\quad 2 \log _{2} 3\left(\log _{9} 2+\log _{9} 4\right)&=\log _{2} 3^{2} \times \log _{9}(2 \times 4)\\\\ &=\log _{2} 9 \times \log _{9} 8\\\\ &=\dfrac{\log 9}{\log 2} \times \frac{\log 8}{\log 9}\\\\ &=\dfrac{\log 8}{\log 2}\\\\ &=\dfrac{\log 2^{3}}{\log 2}\\\\ &=3 \end{aligned}$



5. Compute

(a) $3^{\log _{2} 5}-5^{\log _{2} 3}$

(b) $4^{\log _{2} 3}$

(c) $2^{\log _{2 \sqrt{2}} 27}$





$\begin{aligned} &\text { (a) }\quad 3^{\log _{2} 5}-5^{\log _{2} 3}=3^{\log _{2} 5}-3^{\log _{2} 5}=0\\\\ &\text { (b) }\quad 4^{\log _{2} 3}=3^{\log _{2} 4}=3^{\log _{2} 2^{2}}=3^{2}=9\\\\ &\text { (c) }\quad 2^{\log _{2 \sqrt{2}} 27}=27^{\log _{2 \sqrt{2}} 2}=27^{\log _{2^{\frac{3}{2}}}2}=27^{\frac{2}{3}}=\left(3^{3}\right)^{\frac{2}{3}}=3^{2}=9 \end{aligned}$

Exercise 3.5

1. Given that $\log 2.345=0.3701$. What are the characteristics and the mantissas of each of the followings?

(a) $\log 234,500$  (b) $\log 0.0002345$





$\log 2.345=0.3701$ $\begin{aligned} \text{ (a) }\quad \log 234,500&=\log \left(2.345 \times 10^{5}\right)\\\\ &=\log 2.345+\log 10^{5}\\\\ &=0.3701+5\\\\ &=5+0.3701 \end{aligned}$ characteristics $=5$ mantissa $=0.3701$ $\begin{aligned} \text{ (b) }\quad \log 0.0002345&=\log \left(2.345 \times 10^{-4}\right)\\\\ &=\log 2.345+\log 10^{-4}\\\\ &=0.3701-4\\\\ &=-4+0.3701 \end{aligned}$ characteristics $=-4$ mantissa $=0.3701$



2. Given that $\log_{10}2.74=0.4377$, $\log_{10}2.83=0.4518$, $\log_{10}5.97=0.7760$,
   $\log_{10}6.21=0.7931$, $\log_{10}8.18=0.9128$ and $\log_{10}9.27=0.9671$, compute

(a) $\left(\dfrac{28.3}{597 \times 621}\right)^{2}$

(b) $\dfrac{274^{\frac{1}{3}}}{927 \times 818}$

(c) $\dfrac{28.3 \sqrt{0.621}}{597}$





$\log _{10} 2.74=0.4377, \quad\log _{10} 2.83=0.4518, \quad\log _{10} 5.97=0.7760$, $\log _{10} 6.21=0.7931, \quad \log _{10} 8.18=0.9128, \quad \log _{10} 9.27=0.9671$ (a)  $\left(\dfrac{28.3}{597 \times 621}\right)^{2}$ $\begin{aligned} \text{ Let } p&=\left(\dfrac{28.3}{597 \times 621}\right)^{2}\\\\ \log _{10} p&=\log _{10}\left(\dfrac{28.3}{597 \times 621}\right)^{2}\\\\ &=2\left[\log _{10} 28 \cdot 3-\left(\log _{10} 597+\log _{10} 621\right)\right]\\\\ &=2[0.4518-\log 597-\log 621] \\\\ &=2\left[0.4518-\log \left(5.97 \times 10^{2}\right)-\log _{10}\left(6.21 \times 10^{2}\right)\right] \\\\ &=2\left[0.4518-\log 5.97-\log 10^{2}-\log 6.21-\log 10^{2}\right] \\\\ &=2[0.4518-0.7760-2-0.7931-2] \\\\ &=2(-4.1173) \\\\ &=-8.2346\\\\ \log _{10} p &=-8.2346 \\\\ p&= 10^{-8.2346} \end{aligned}$ $\therefore \left(\dfrac{28.3}{597 \times 621}\right)^{2}=10^{-8.2346}$ (b)  $\dfrac{274^{\frac{1}{3}}}{927 \times 818}$ $ \begin{aligned} \text { Let } p &=\dfrac{274^{\frac{1}{3}}}{927 \times 818} \\\\ \log _{10} p &=\log _{10}\left(\dfrac{274^{\frac{1}{3}}}{927 \times 818}\right) \\\\ &=\log _{10} 274^{\frac{1}{3}}-\log _{10}(927 \times 818) \\\\ &=\dfrac{1}{3} \log _{10}(274)-\log _{10^{9}} 927-\log _{10} 818 \\\\ &=\dfrac{1}{3} \log _{10}\left(2.74 \times 10^{2}\right)-\log _{10}\left(9.27 \times 10^{2}\right)-\log _{10}(8.18) \\\\ &=\dfrac{1}{3}\left[\log _{10} 2.74+\log _{10} 10^{2}\right]-\log _{10} 9.27-\log _{10} 10^{2}-\log _{10} 8.18 \\\\ &=\dfrac{1}{3}(0.4378+2)-0.9671-2-0.9128-2 \\\\ &=\dfrac{1}{3}(2.4378)-5.8899 \\\\ &=0.8126-5.8799 \\\\ \log _{10} p &=-5.0673 \\\\ p&= 10^{-5.0673} \end{aligned}$ $\therefore \dfrac{274^{\frac{1}{3}}}{927 \times 818}=10^{-5.0673}$ (c) $\dfrac{28.3 \sqrt{0.621}}{597}$ $\begin{aligned} \text { Let } p&=\dfrac{28.3 \sqrt{0.621}}{597}\\\\ \log _{10} p&=\log _{10} \dfrac{28.3 \sqrt{0.621}}{597}\\\\ &=\log _{10} 28 \cdot 3+\log _{10} \sqrt{0.621}-\log _{10} 597\\\\ &=\log _{10}\left(2.83 \times 10^{1}\right)+\log _{10}(0.621)^{\frac{1}{2}}-\log _{10}\left(5.97 \times 10^{2}\right)\\\\ &=\log _{10} 2.83+\log _{10} 10+\dfrac{1}{2} \log _{10}\left(6.21 \times 10^{-1}\right)-\log _{10} 5.97-\log _{10} 10^{2}\\\\ &=0.4518+1+\dfrac{1}{2}\left(\log _{10} 6 \cdot 21+\log _{10} 10^{-1}\right)-0.7760-2\\\\ &=-1.3243+\dfrac{1}{2}(0.7931-1)\\\\ &=-1.3243+\frac{1}{2}(-0.2069)\\ &=-1.3243-0.10345\\\\ \log _{10} p&=-1.4278\\\\ p&=10^{-1.4278} \end{aligned}$ $\therefore \dfrac{28.3 \sqrt{0.621}}{597}=10^{-1.4278}$