Logarithm: Miscellaneous Exercise

Definition: Logarithm

Let $N$ and $b$ be positive real numbers, with $b \neq 1$. Then the logarithm of $N$ (with respect) to the base $b$ is the exponent by which $b$ must be raised to yield $N$, and is denoted by $\log _{b} N$

Rules of Logarithms

$\begin{array}{ll} \text{L}1. & N=b^{\log _{b} N}\\\\ \text{L}2. & x=\log _{b} b^{x}\\\\ \text{L}3. & \log _{b} b=1\\\\ \text{L}4. & \log _{b} 1=0\\\\ \text{L}5. & \log _{b}(M N)=\log _{b} M+\log _{b} N\\\\ \text{L}6. & \log _{b} N^{p}=p \log _{b} N\\\\ \text{L}7. & \log _{b}\left(\displaystyle\frac{M}{N}\right)=\log _{b} M-\log _{b} N\\\\ \text{L}8. & \log _{a} N=\displaystyle\frac{\log _{b} N}{\log _{b} N}\\\\ \text{L}9. & \log _{a} N=\displaystyle\frac{1}{\log _{N} a}\\\\ \text{L}10. & \log _{a^{p}} N=\displaystyle\frac{1}{p} \log _{a} N\\\\ \text{L}11. & a^{\log _{k} b}=b^{\log _{k} a} \end{array}$

Common Logarithm

The logarithm of $N$ to the base $10\left(\log _{10} N\right)$ is said to be a common logarithm, and is usually written as $\log N$ (omitting the base). where $n$ is called the characteristic and $\log a$ is called the mantissa of $\log N$.

$\begin{array}{|l|} \hline\log _{10} N=\log N\\ \hline \end{array}$

If $\quad N=a \times 10^{n}$,

then $\quad \log N=\log \left(a \times 10^{n}\right)=\log 10^{n}+\log a=n+\log a$ where $n$ is called the characteristic and $\log a$ is called the mantissa of $\log N$.

Note that $n$ is an integer and $1 \leq a<10$.

Euler's Number

As a positive integer $n$ become very large, the value of $\left(1+\displaystyle\frac{1}{n}\right)^{n}$ approaches an irrational number, which is denoted by $e$.

Natural Logarithm

The logarithm of $N$ to the base $e$ is called a natural logarithm, and is denoted by $\ln N$.

$\begin{array}{|l|} \hline\log _{e} N=\ln N\\ \hline \end{array}$

1. Write the following expressions in terms of $\log x$, $\log y$ and $\log z$.

(a)	$\log (x^{2} y)$		Show Solution
(b)	$\log \displaystyle\frac{x^{3} y^{2}}{z}$		Show Solution
(c)	$\log \displaystyle\frac{\sqrt{x} \sqrt[3]{y^{2}}}{z^{4}}$		Show Solution
(d)	$\log (x y z)$		Show Solution
(e)	$\log \left(\displaystyle\frac{x}{y z}\right)$		Show Solution
(f)	$\log \left(\displaystyle\frac{x}{y}\right)^{2}$		Show Solution
(g)	$\log \left(x y\right)^{\frac{1}{3}}$		Show Solution
(h)	$\log (x \sqrt{z})$		Show Solution
(i)	$\log \displaystyle\frac{\sqrt[3]{x}}{\sqrt[3]{y z}}$		Show Solution
(j)	$\log \sqrt[4]{\displaystyle\frac{x^{3} y^{2}}{z^{4}}}$		Show Solution
(k)	$\log \left(x \sqrt{\displaystyle\frac{\sqrt{x}}{z}}\right)$		Show Solution
(l)	$\log \sqrt{\displaystyle\frac{x y^{2}}{z^{8}}}$		Show Solution



2. Prove the following statements.

(a)	$\quad\log _{\sqrt{b}} x=2 \log _{b} x$		Show Solution
(b)	$\quad\log _{\frac{1}{\sqrt{b}}} \sqrt{x}=-\log _{b} x$		Show Solution
(c)	$\quad\log _{b^{4}} x^{2}=\log _{b} \sqrt{x}$		Show Solution



3. Given that $\log 2=x, \log 3=y$ and $\log 7=z$, express the following expressions in terms of $x, y$, and $z$.

(a)	$\log 12$		Show Solution
(b)	$\log 200$		Show Solution
(c)	$\log \displaystyle\frac{14}{3}$		Show Solution
(d)	$\log 0.3$		Show Solution
(e)	$\log 1.5$		Show Solution
(f)	$\log 10.5$		Show Solution
(g)	$\log 15$		Show Solution
(h)	$\log \displaystyle\frac{6000}{7}$		Show Solution



4. Solve the following logarithmic equations.



(a)	$\ln x=-3$		Show Solution
(b)	$\log (3 x-2)=2$		Show Solution
(c)	$2 \log x=\log 2+\log (3 x-4)$		Show Solution
(d)	$\log x+\log (x-1)=\log (4 x)$		Show Solution
(e)	$\log _{3}(x+25)-\log _{3}(x-1)=3$		Show Solution
(f)	$\log _{9}(x-5)+\log _{9}(x+3)=1$		Show Solution
(g)	$\log x+\log (x-3)=1$		Show Solution
(h)	$\log _{2}(x-2)+\log _{2}(x+1)=2$		Show Solution



5. Find the inverse of each of the following functions.

(a) $\quad f(x)=\log _{2}(x-3)-5$

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(b) $\quad f(x)=3 \log _{3}(x+3)+1$

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(c) $\quad f(x)=-2 \log 2(x-1)+2$

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(d) $\quad f(x)=-\ln (1-2 x)+1$

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(e) $\quad f(x)=2^{x}-3$

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(f) $\quad f(x)=2 \cdot 3^{3 x}-1$

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(g) $\quad f(x)=-5 \cdot e^{-x}+2$

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(h) $\quad f(x)=1-2 e^{-2 x}$

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Logarithms : Exercise (3.5) - Solutions

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) $\log 0.0002345$

Show/Hide Solution

$\begin{array}{l}\ \ \ \ \ \ \ \ \log 2.345=0.3701\\\\\text{(a)}\ \ \ \log 234,500\\\\\ \ \ =\log \left( {2.345\times {{{10}}^{5}}} \right)\\\\\ \ \ =\log {{10}^{5}}+\log 2.345\\\\\ \ \ =5+0.3701\\\\\ \ \ =5.3701\\\\\therefore \ \text{characteristic}=5\\\\\ \ \ \text{mantissa}=.3701\\\\\\\text{(b)}\ \ \ \log 0.0002345\\\\\ \ \ =\log \left( {2.345\times {{{10}}^{{-4}}}} \right)\\\\\ \ \ =\log {{10}^{{-4}}}+\log 2.345\\\\\ \ \ =-4+0.3701\\\\\therefore \ \text{characteristic}=-4\\\\\ \ \ \text{mantissa}=.3701\end{array}$

2.          Using $\log _{10} 2.74=0.4377, \log _{10} 2.83=0.4518, \log _{10} 5.97=0.7760$, $\log _{10} 6.21=0.7931, \log _{10} 8.18=0.9128$ and $\log _{10} 9.27=0.9671$, compute

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

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

(c) $\displaystyle\frac{28.3 \sqrt{0.621}}{597}$

Show/Hide Solution

$\begin{array}{l}\ \ \ \ \ {{\log }_{{10}}}2.74=0.4377,\\\ \ \ \ \ {{\log }_{{10}}}2.83=0.4518,\\\ \ \ \ \ {{\log }_{{10}}}5.97=0.7760,\\\ \ \ \ \ {{\log }_{{10}}}6.21=0.7931,\\\ \ \ \ \ {{\log }_{{10}}}8.18=0.9128,\\\ \ \ \ \ {{\log }_{{10}}}9.27=0.9671.\\\text{(a)}\ \ \text{Let}\ x={{\left( {\displaystyle\frac{{28.3}}{{597\times 621}}} \right)}^{2}},\ \text{then}\\\ \ \ \ \ \log x=\log {{\left( {\displaystyle\frac{{28.3}}{{597\times 621}}} \right)}^{2}}\\\ \ \ \ \ \ \ \ \ \ \ \ =2\log \left( {\displaystyle\frac{{28.3}}{{597\times 621}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =2\left( {\log 28.3-\log (597\times 621)} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =2\left( {\log 28.3-\log 597-\log 621} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =2\left( {\log 28.3-\log 597-\log 621} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =2\left( {1.4518-2.7760-2.7931} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =-8.2346\\\ \ \ \ \ x={{10}^{{-8.2346}}}\\\\\text{(b)}\ \ \text{Let}\ x=\displaystyle\frac{{{{{274}}^{{\displaystyle\frac{1}{3}}}}}}{{927\times 818}},\ \text{then}\\\ \ \ \ \ \log x=\log \left( {\displaystyle\frac{{{{{274}}^{{\displaystyle\frac{1}{3}}}}}}{{927\times 818}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{1}{3}\log 274-\log \left( {927\times 818} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{1}{3}\log 274-\left( {\log 927+\log 818} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{1}{3}\left( {2.4377} \right)-2.9671-2.9128\\\ \ \ \ \ \ \ \ \ \ \ \ =-5.0673\\\ \ \ \ \ x={{10}^{{-5.0673}}}\\\\\text{(c)}\ \ \text{Let}\ x=\displaystyle\frac{{28.3\sqrt{{0.621}}}}{{597}},\ \text{then}\\\ \ \ \ \ \log x=\log \left( {\displaystyle\frac{{28.3\sqrt{{0.621}}}}{{597}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =\log 28.3+\displaystyle\frac{1}{2}\log 0.621-\log 597\\\ \ \ \ \ \ \ \ \ \ \ \ =1.4518+\displaystyle\frac{1}{2}\left( {-1+0.7931} \right)-2.7760\\\ \ \ \ \ \ \ \ \ \ \ \ =-1.4277\\\ \ \ \ \ x={{10}^{{-1.4277}}}\end{array}$

3.          Express each to a natural logarithm in terms of $\ln 2$ and $\ln 5$.

(a) $\log_{8}25$

(b) $\log_{5}64$

(c) $\log_{25}100$

Show/Hide Solution

$\begin{array}{l}\text{(a)}\ \ \ \ {{\log }_{8}}25\ \\\ \ \ \ =\displaystyle\frac{{\ln 25}}{{\ln 8}}\\\ \ \ \ =\displaystyle\frac{{\ln {{5}^{2}}}}{{\ln {{2}^{3}}}}\\\ \ \ \ =\displaystyle\frac{{2\ln 5}}{{3\ln 2}}\\\\\text{(b)}\ \ \ \ {{\log }_{5}}64\ \\\ \ \ \ =\displaystyle\frac{{\ln 64}}{{\ln 5}}\\\ \ \ \ =\displaystyle\frac{{\ln {{2}^{6}}}}{{\ln 5}}\\\ \ \ \ =\displaystyle\frac{{6\ln 5}}{{\ln 2}}\\\\\text{(c)}\ \ \ \ {{\log }_{{25}}}100\ \\\ \ \ \ =\displaystyle\frac{{\ln 100}}{{\ln 25}}\\\ \ \ \ =\displaystyle\frac{{\ln ({{5}^{2}}\times {{2}^{2}})}}{{\ln {{5}^{2}}}}\\\ \ \ \ =\displaystyle\frac{{2\ln 5+2\ln 2}}{{2\ln 5}}\\\ \ =1+\displaystyle\frac{{\ln 2}}{{\ln 5}}\ \ \ \end{array}$

4.          Write each expression as a single logarithm.

(a) $3\ln \left( {t + 5} \right) - 4\ln t - 2\ln \left( {s - 1} \right)$

(b) $2\ln x + 5\ln y - \displaystyle\frac{1}{2}\ln z$

(c) $\displaystyle \frac{1}{3}\ln a - 6\ln b + 2$

Show/Hide Solution

$\begin{array}{l} \text{(a)}\ \ \ \ 3\ln \left( {t+5} \right)-4\ln t-2\ln \left( {s-1} \right)\\\\ \ \ \ \ =\ln {{\left( {t+5} \right)}^{3}}-\ln {{t}^{4}}-\ln {{\left( {s-1} \right)}^{2}}\\\\ \ \ \ \ =\ln \displaystyle\frac{{{{{\left( {t+5} \right)}}^{3}}}}{{{{t}^{4}}{{{\left( {s-1} \right)}}^{2}}}}\\\\\\ \text{(b)}\ \ \ \ 2\ln x+5\ln y-\displaystyle\frac{1}{2}\ln z\\\\ \ \ \ \ =\ln {{x}^{2}}+\ln {{y}^{5}}-\ln \sqrt{z}\\\\ \ \ \ \ =\ln \displaystyle\frac{{{{x}^{2}}\cdot {{y}^{5}}}}{{\sqrt{z}}}\\\\\\ \text{(c)}\ \ \ \ \displaystyle\frac{1}{3}\ln a-6\ln b+2\\\\ \ \ \ \ =\ln {{a}^{{\frac{1}{3}}}}-\ln {{b}^{6}}+\ln {{e}^{2}}\\\\ \ \ \ \ =\ln \displaystyle\frac{{\sqrt[3]{a}\cdot {{e}^{2}}}}{{{{b}^{6}}}} \end{array}$

5.          Solve the following logarithmic equations for $x$.

(a) $\ln \left( x \right) + \ln \left( {x + 3} \right) = \ln \left( {20 - 5x} \right)$

(b) $2\ln \left( {\sqrt x } \right) - \ln \left( {1 - x} \right) = 2$

Show/Hide Solution

$\begin{array}{l}\text{(a)}\ \ \ln (x)+\ln \left( {x+3} \right)=\ln \left( {20-5x} \right)\\\ \\\ \ \ \ \ \ \ln \left( {x\left( {x+3} \right)} \right)=\ln \left( {20-5x} \right)\\\\\ \ \ \ \ \ x\left( {x+3} \right)=20-5x\\\\\ \ \ \ \ \ {{x}^{2}}+8x-20=0\\\\\ \ \ \ \ \ (x+10)(x-2)=0\\\\\ \ \ \ \ \ x=-10\ \text{or}\ x=2\\\\\ \ \ \ \ \ \text{Since }x>0,\ x=-10\ \text{is impossible}\text{.}\\\\\ \ \ \therefore \ \ x=2\\\\\ \ \ \ \ \\\text{(b)}\ \ \ 2\ln \left( {\sqrt{x}} \right)-\ln \left( {1-x} \right)=2\\\ \\\ \ \ \ \ \ln \left( {\displaystyle\frac{{{{{\left( {\sqrt{x}} \right)}}^{2}}}}{{1-x}}} \right)=2\\\\\ \ \ \ \ \displaystyle\frac{x}{{1-x}}={{e}^{2}}\\\ \\\ \ \ \ \ x={{e}^{2}}-{{e}^{2}}x\\\\\ \ \ \ \ {{e}^{2}}x+x={{e}^{2}}\\\\\ \ \ \ \ x\left( {{{e}^{2}}+1} \right)={{e}^{2}}\\\\\ \ \ \ \ x=\displaystyle\frac{{{{e}^{2}}}}{{{{e}^{2}}+1}}\end{array}$

6.          Solve the equations:

(a) $x - xe^{5x + 2} = 0$

(b) $7 + 15e^{1 - 3x} = 10$ using $\ln 5 = 1.6094$.

(c) $4e^{1 + 3x} - 9e^{5 - 2x} = 0$ using $\ln 2=0.6931$ and $\ln 3=1.0986$.

No(3) မှ No(6) အထိ မေးခွန်းများသည် ပြဌာန်းစာအုပ်တွင် မပါဝင်ပါ Natural Logarithm ဆိုင်ရာ ပုစ္တာများကို တိုးချဲ့လေ့ကျင့်နိုင်ရန် ပေါင်းထည့်ပေးထားခြင်း ဖြစ်ပါသည်။

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$\begin{array}{l}\text{(a)}\ \ \ x-x{{e}^{{5x+2}}}=0\\\ \\\ \ \ \ \ \ x\left( {1-{{e}^{{5x+2}}}} \right)=0\\\\\ \ \ \ \ \ x=0\ \text{or}\ 1-{{e}^{{5x+2}}}=0\\\\\ \ \ \ \ \ x=0\ \text{or}\ {{e}^{{5x+2}}}=1\\\\\ \ \ \ \ \ x=0\ \text{or}\ 5x+2=\ln 1\\\\\ \ \ \ \ \ x=0\ \text{or}\ 5x+2=0\\\\\ \ \therefore \ \ \ x=0\ \text{or}\ x=-\displaystyle\frac{2}{5}\\\\\ \ \ \ \ \\\text{(b)}\ \ \ 7+15{{e}^{{1-3x}}}=10\\\ \\\ \ \ \ \ \ 15{{e}^{{1-3x}}}=3\\\\\ \ \ \ \ \ {{e}^{{1-3x}}}=\displaystyle\frac{1}{5}\\\ \\\ \ \ \ \ \ 1-3x=\ln \left( {\displaystyle\frac{1}{5}} \right)\\\\\ \ \ \ \ \ 1-3x=\ln \left( {{{5}^{{-1}}}} \right)\\\\\ \ \ \ \ \ 1-3x=-\ln \left( 5 \right)\\\\\ \ \ \ \ \ 3x=1+\ln \left( 5 \right)\\\\\ \ \ \ \ \ 3x=1+\text{1}\text{.6094}\\\\\ \ \ \ \ \ 3x=2.\text{6094}\\\\\ \ \ \ \ \ x=\text{0}\text{.8698}\\\\\\\text{(c)}\ \ \ 4{{e}^{{1+3x}}}-9{{e}^{{5-2x}}}=0\\\\\ \ \ \ \ 4{{e}^{{1+3x}}}=9{{e}^{{5-2x}}}\\\\\ \ \ \ \ \displaystyle\frac{{{{e}^{{1+3x}}}}}{{{{e}^{{5-2x}}}}}=\displaystyle\frac{9}{4}\\\\\ \ \ \ \ {{e}^{{5x-4}}}=\displaystyle\frac{9}{4}\\\\\ \ \ \ \ 5x-4=\ln \displaystyle\frac{{{{3}^{2}}}}{{{{2}^{2}}}}\\\\\ \ \ \ \ 5x=2\left( {\ln 3-\ln 2} \right)+4\\\\\ \ \ \ \ x=\displaystyle\frac{2}{5}\left( {\ln 3-\ln 2+2} \right)\\\\\ \ \ \ \ x=\displaystyle\frac{2}{5}\left( {\text{1}\text{.0986}-\text{0}\text{.6931+2}} \right)\\\\\ \ \ \ \ x=\text{0}\text{.9622}\end{array}$