Logarithms : Exercise (3.3) Solutions

$\begin{array}{|l|l|l|} \hline {\ \ \ \ \ \ \ \ \ \ \text { Properties }} & {\ \ \ \ \ \ \ \text { For Exponents }} & {\ \ \ \ \ \ \ \ \ \ \ \ \ \text { For Logarithms }} \\ \hline \text { One-to-one Property } & \text { If } b^{x}=b^{y}, \text { then } x=y & \text { If } \log _{b} M=\log _{b} N, \text { then } M=N \\ \hline \text { Product Property } & b^{x} \cdot b^{y}=b^{x+y} & \log _{b}(M N)=\log _{b} M+\log _{b} N \\ \hline \text { Quotient Property } & \displaystyle\frac{b^{x}}{b^{y}}=b^{x-y} & \log _{b} \displaystyle\frac{M}{N}=\log _{b} M-\log _{b} N \\ \hline \text { Power Property } & \left(b^{x}\right)^{y}=b^{x y} & \log _{b} N^{p}=p \log _{b} N \\ \hline \end{array}$

1.           Replace $\square$ with the appropriate number.

              $\begin{array}{l} \text{(a)}\ \ \log _{3} 24=\log _{3} 6+\log _{3} \square\\ \text{(b)}\ \ \log _{5} 24=\log _{5} 60+\log _{5} \square\\ \text{(c)}\ \ \log _{2} \square=3 \log _{2} 3\\ \text{(d)}\ \ \log _{10} 9=\square \log _{10} 3\\ \text{(e)}\ \ \log _{8} 5=\log _{8} \square-\log _{8} 11 \end{array}$

Show/Hide Solution

$\begin{array}{l} \text{(a)}\ \ 4\\ \text{(b)}\ \ \displaystyle\frac{2}{5}\\ \text{(c)}\ \ 27\\ \text{(d)}\ \ 2\\ \text{(e)}\ \ 55 \end{array}$

2.           Write each expression as a single logarithm.

              $\begin{array}{l} \text{(a)}\ \ \log _{b} 20+\log _{b} 57-\log _{b} 241\\ \text{(b)}\ \ 3 \log _{b} 8-\displaystyle\frac{1}{2} \log _{b} 12\\ \text{(c)}\ \ \log _{b} x-2 \log _{b} y-\log _{b} a\\ \text{(d)}\ \ \log _{2} 3+\log _{4} 15 \end{array}$

Show/Hide Solution

$\begin{array}{l}\text{(a)}\;\;\ \ {{\log }_{b}}20+{{\log }_{b}}57-{{\log }_{b}}241\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{20\times 57}}{{241}}\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{1140}}{{241}}\\\\ \text{(b)}\;\;\ \ 3{{\log }_{b}}8-\displaystyle \frac{1}{2}{{\log }_{b}}12\\ \ \ \ \ =\ {{\log }_{b}}{{8}^{3}}-{{\log }_{b}}{{12}^{{\displaystyle \frac{1}{2}}}}\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{8\times 8\times 8}}{{\sqrt{{12}}}}\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{8\times 8\times 8}}{{2\sqrt{3}}}\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{256}}{{\sqrt{3}}}\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{256\sqrt{3}}}{3}\\\\ \text{(c)}\;\;\ \ {{\log }_{b}}x-2{{\log }_{b}}y-{{\log }_{b}}a\\ \ \ \ \ =\ {{\log }_{b}}x-{{\log }_{b}}{{y}^{2}}-{{\log }_{b}}a\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{x}{{a{{y}^{2}}}}\\\\ \text{(d)}\;\;\ \ {{\log }_{2}}3+{{\log }_{4}}15\\ \ \ \ \ \text{Let}\ {{\log }_{4}}15=x,\ \text{then}\\ \ \ \ \ 15={{4}^{x}}\\\ \ \ \ 15={{2}^{2}}^{x}\\ \ \ \ \ \therefore \ \ 2x={{\log }_{2}}15\\ \ \ \ \ \therefore \ \ x=\displaystyle \frac{1}{2}{{\log }_{2}}15\\ \ \ \ \ \therefore \ \ {{\log }_{4}}15={{\log }_{2}}\sqrt{{15}}\\ \ \ \ \ \therefore \ \ {{\log }_{2}}3+{{\log }_{4}}15\\ \ \ \ \ =\ \ {{\log }_{2}}3+{{\log }_{2}}\sqrt{{15}}\\ \ \ \ \ =\ \ {{\log }_{2}}3\sqrt{{15}} \end{array}$

3.           Write each expression in terms of $\log _{b} 2, \log _{b} 3$ and $\log _{b} 5$.

              $\begin{array}{l} \text{(a)}\ \ \log _{b} 8\\ \text{(b)}\ \ \log _{b} 15\\ \text{(c)}\ \ \log _{b} 270\\ \text{(d)}\ \ \log _{b} \displaystyle\frac{27 \sqrt[3]{5}}{16}\\ \text{(e)}\ \ \log _{b} \displaystyle\frac{216}{\sqrt[3]{32}}\\ \text{(f)}\ \ \log _{b}(648 \sqrt{125})\\ \end{array}$

Show/Hide Solution

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

4.           Evaluate each expression.

              $\begin{array}{l} \text{(a)}\ \ \log _{2} 128\\ \text{(b)}\ \ \log _{3} 81^{4}\\ \text{(c)}\ \ \log _{\frac{1}{2}} 8\\ \text{(d)}\ \ \log _{8} 2\\ \text{(e)}\ \ \log _{3} \displaystyle\frac{\sqrt{3}}{81}\\ \text{(f)}\ \ \displaystyle\frac{\log _{3} \sqrt{3}}{\log _{3} 81}\\ \text{(g)}\ \ \displaystyle\frac{\log _{2} 25}{\log _{2} 5}\\ \text{(h)}\ \ \log _{4} 8 \end{array}$

Show/Hide Solution

$\begin{array}{l} \text{(a)}\;\;{{\log }_{2}}128={{\log }_{2}}{{2}^{7}}=7\\\\ \text{(b)}\;\;{{\log }_{3}}{{81}^{4}}={{\log }_{3}}{{\left( {{{3}^{4}}} \right)}^{4}}={{\log }_{3}}{{3}^{{16}}}=16\\\\ \text{(c)}\;\;{{\log }_{{\frac{1}{2}}}}8={{\log }_{{\frac{1}{2}}}}{{2}^{3}}={{\log }_{{\frac{1}{2}}}}{{\left( {\displaystyle\frac{1}{2}} \right)}^{{-3}}}=-3\\\\ \text{(d)}\;\;{{\log }_{8}}2={{\log }_{8}}{{8}^{{\frac{1}{3}}}}=\displaystyle \frac{1}{3}\\\\ \text{(e)}\;\;{{\log }_{3}}\displaystyle \frac{{\sqrt{3}}}{{81}}={{\log }_{3}}\displaystyle \frac{{{{3}^{{\frac{1}{2}}}}}}{{{{3}^{4}}}}={{\log }_{3}}{{3}^{{-\frac{7}{2}}}}=-\displaystyle \frac{7}{2}\\\\ \text{(f)}\;\;\displaystyle \frac{{{{{\log }}_{3}}\sqrt{3}}}{{{{{\log }}_{3}}81}}=\displaystyle \frac{{{{{\log }}_{3}}{{3}^{{\frac{1}{2}}}}}}{{{{{\log }}_{3}}{{3}^{4}}}}=\displaystyle \frac{{\frac{1}{2}}}{4}=\displaystyle \frac{1}{8}\\\\ \text{(g)}\;\;\displaystyle \frac{{{{{\log }}_{2}}25}}{{{{{\log }}_{2}}5}}=\displaystyle \frac{{{{{\log }}_{2}}{{5}^{2}}}}{{{{{\log }}_{2}}5}}=\displaystyle \frac{{2{{{\log }}_{2}}5}}{{{{{\log }}_{2}}5}}=2\\\\ \text{(h)}\;\;{{\log }_{4}}8={{\log }_{4}}\sqrt{{64}}={{\log }_{4}}{{4}^{{\frac{3}{2}}}}=\displaystyle \frac{3}{2} \end{array}$

5.           Use $\log _{10} 2=0.3010$ and $\log _{10} 3=0.4771$ to evaluate each of the following expressions.

              $\begin{array}{lll} \text{(a)}\ \ \log _{10} 6 & \text{(b)}\ \ \log _{10} 1.5 & \text{(c)}\ \ \log _{10} \sqrt{3}\\\\ \text{(d)}\ \ \log _{10} 4 & \text{(e)}\ \ \log _{10} 4.5 & \text{(f)}\ \ \log _{10} 8\\\\ \text{(g)}\ \ \log _{10} 18 & \text{(h)}\ \ \log _{10} 5 \end{array}$

Show/Hide Solution

$\begin{array}{l} {{\log }_{{10}}}2=0.3010,\ {{\log }_{{10}}}3=0.4771\\\\ \text{(a)}\;\;{{\log }_{{10}}}6=\;{{\log }_{{10}}}\left( {2\times 3} \right)\\ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{{10}}}2+{{\log }_{{10}}}3\\ \ \ \ \ \ \ \ \ \ \ \ \ =0.3010+0.4771\\\ \ \ \ \ \ \ \ \ \ \ \ =0.7781\\\\ \text{(b)}\;\;{{\log }_{{10}}}1.5=\;{{\log }_{{10}}}\displaystyle \frac{3}{2}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;{{\log }_{{10}}}3-{{\log }_{{10}}}2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.4771-0.3010\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.1761\\\\ \text{(c)}\;\;{{\log }_{{10}}}\sqrt{3}\ =\;{{\log }_{{10}}}{{3}^{{\frac{1}{2}}}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;\displaystyle \frac{1}{2}{{\log }_{{10}}}3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;\displaystyle \frac{1}{2}\times 0.4771\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;0.2386\\\\\text{(d)}\;\;{{\log }_{{10}}}4={{\log }_{{10}}}{{2}^{2}}\\ \ \ \ \ \ \ \ \ \ \ \ \ =2{{\log }_{{10}}}2\\ \ \ \ \ \ \ \ \ \ \ \ \ =2\ (0.3010)\\ \ \ \ \ \ \ \ \ \ \ \ \ =0.6020\\\\ \text{(e)}\;\;{{\log }_{{10}}}4.5=\;{{\log }_{{10}}}\displaystyle \frac{9}{2}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;{{\log }_{{10}}}\displaystyle \frac{{{{3}^{2}}}}{2}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;2{{\log }_{{10}}}3-{{\log }_{{10}}}2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;2\left( {0.4771} \right)-0.3010\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;0.6532\\\\ \text{(f)}\;\;{{\log }_{{10}}}8\ ={{\log }_{{10}}}{{2}^{3}}\\ \ \ \ \ \ \ \ \ \ \ \ \ =3{{\log }_{{10}}}2\\ \ \ \ \ \ \ \ \ \ \ \ \ =3\left( {0.3010} \right)\\ \ \ \ \ \ \ \ \ \ \ \ \ =0.9030\\\\ \text{(g)}\;\;{{\log }_{{10}}}18={{\log }_{{10}}}\left( {2\times {{3}^{2}}} \right)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;{{\log }_{{10}}}2+2{{\log }_{{10}}}3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.3010+2\left( {0.4771} \right)\\\\ \text{(h)}\;\;{{\log }_{{10}}}5={{\log }_{{10}}}\displaystyle \frac{{10}}{2}\\ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{{10}}}10-{{\log }_{{10}}}2\\ \ \ \ \ \ \ \ \ \ \ \ \ =1-0.3010\\ \ \ \ \ \ \ \ \ \ \ \ \ =0.6990 \end{array}$

6.           Solve the following equations for $x$.

              $\begin{array}{lll} \text{(a)}\ \ \log _{a} \displaystyle\frac{18}{5}+\log _{a} \displaystyle\frac{10}{3}-\log _{a} \displaystyle\frac{6}{7}=\log _{a} x\\\\ \text{(b)}\ \ \log _{b} x=2-a+\log _{b}\left(\displaystyle\frac{a^{2} b^{a}}{b^{2}}\right)\\\\ \text{(c)}\ \ \log x^{3}-\log x^{2}=\log 5 x-\log 4 x\\\\ \text{(d)}\ \ \log _{10} x+\log _{10} 3=\log _{10} 6\\\\ \text{(e)}\ \ 8 \log x=\log a^{\frac{3}{2}}+\log 2-\displaystyle\frac{1}{2} \log a^{3}-\log \frac{2}{a^{4}}\\\\ \end{array}$

Show/Hide Solution

$\begin{array}{l} \text{(a)}\;\;{{\log }_{a}}\displaystyle \frac{{18}}{5}+{{\log }_{a}}\displaystyle \frac{{10}}{3}-{{\log }_{a}}\displaystyle \frac{6}{7}={{\log }_{a}}x\\ \ \ \ \ {{\log }_{a}}\left( {\displaystyle \frac{{\displaystyle \frac{{18}}{5}\times \displaystyle \frac{{10}}{3}}}{{\displaystyle \frac{6}{7}}}} \right)={{\log }_{a}}x\\ \ \ \ \ x=\displaystyle \frac{{18}}{5}\times \displaystyle \frac{{10}}{3}\times \displaystyle \frac{7}{6}\\ \ \ \ \ x=14\\\\ \text{(b)}\;\;{{\log }_{b}}x=2-a+{{\log }_{b}}\left( {\displaystyle \frac{{{{a}^{2}}{{b}^{a}}}}{{{{b}^{2}}}}} \right)\\ \ \ \ \ {{\log }_{b}}x=2-a+{{\log }_{b}}{{a}^{2}}+{{\log }_{b}}{{b}^{a}}-{{\log }_{b}}{{b}^{2}}\\ \ \ \ \ {{\log }_{b}}x=2-a+{{\log }_{b}}{{a}^{2}}+a-2\\ \ \ \ \ {{\log }_{b}}x={{\log }_{b}}{{a}^{2}}\\ \ \ \ \ x={{a}^{2}}\\\\ \text{(c)}\;\;\log {{x}^{3}}-\log {{x}^{2}}=\log 5x-\log 4x\\ \ \ \ \ \log \displaystyle \frac{{{{x}^{3}}}}{{{{x}^{2}}}}=\log \displaystyle \frac{{5x}}{{4x}}\\ \ \ \ \ \log x=\log \displaystyle \frac{5}{4}\\ \ \ \ \ x=\displaystyle \frac{5}{4}\\\\ \text{(d)}\;\;{{\log }_{{10}}}x+{{\log }_{{10}}}3={{\log }_{{10}}}6\\ \ \ \ \ \ {{\log }_{{10}}}3x={{\log }_{{10}}}6\\ \ \ \ \ \ 3x=6\\ \ \ \ \ \ x=2\\ \text{(e)}\;\;8\log x=\log {{a}^{{\frac{3}{2}}}}+\log 2- \frac{1}{2}\log {{a}^{3}}-\log \displaystyle \frac{2}{{{{a}^{4}}}}\\ \ \ \ \ \log {{x}^{8}}=\log {{a}^{{\frac{3}{2}}}}+\log 2-\log {{a}^{{ \frac{3}{2}}}}-\left( {\log 2-\log {{a}^{4}}} \right)\\ \ \ \ \ \log {{x}^{8}}=\log {{a}^{4}}\\ \ \ \ \ {{x}^{8}}={{a}^{4}}\\ \ \ \ \ x=\sqrt{a}\end{array}$

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

Show/Hide Solution

$\begin{array}{l} \ \ \;\;{{\log }_{{10}}}5=0.6990\\\\ \ \ \ \ {{\log }_{{10}}}x=0.2330\\\\ \ \ \ \ \therefore \ \ 3{{\log }_{{10}}}x=0.6990\\\\ \ \ \ \ \therefore \ \ 3{{\log }_{{10}}}x={{\log }_{{10}}}5\\\\ \ \ \ \ \therefore \ \ {{\log }_{{10}}}x=\displaystyle\frac{1}{3}{{\log }_{{10}}}5\\\\ \ \ \ \ \therefore \ \ {{\log }_{{10}}}x={{\log }_{{10}}}{{5}^{{\frac{1}{3}}}}\\\\ \ \ \ \ \therefore \ \ x={{5}^{{\frac{1}{3}}}}=\sqrt[3]{5} \end{array}$

8.           Show that if $\log _{e} I=-\displaystyle\frac{R}{L} t+\log _{e} I_{0}$ then $I=I_{0} e^{-\frac{R t}{L}}$

Show/Hide Solution

$\begin{array}{l} {{\log }_{e}}I=-\displaystyle \frac{R}{L}t+{{\log }_{e}}{{I}_{0}}\\\\ {{\log }_{e}}I-{{\log }_{e}}{{I}_{0}}=-\displaystyle \frac{{Rt}}{L}\\\\ {{\log }_{e}}\displaystyle \frac{I}{{{{I}_{0}}}}=-\displaystyle \frac{{Rt}}{L}\\\\ \displaystyle \frac{I}{{{{I}_{0}}}}={{e}^{{-\frac{{Rt}}{L}}}}\\\\ \therefore \ I={{I}_{0}}{{e}^{{-\frac{{Rt}}{L}}}} \end{array}$

9.           Show that if $\log _{b} y=\displaystyle\frac{1}{2} \log _{b} x+c$ then $y=b^{c} \sqrt{x}$

Show/Hide Solution

$\begin{array}{l}{{\log }_{b}}y=\displaystyle \frac{1}{2}{{\log }_{b}}x+c\\\\ {{\log }_{b}}y-\displaystyle \frac{1}{2}{{\log }_{b}}x=c\\\\ {{\log }_{b}}y-{{\log }_{b}}{{x}^{{\frac{1}{2}}}}=c\\\\ {{\log }_{b}}\displaystyle \frac{y}{{\sqrt{x}}}=c\\\\ \displaystyle \frac{y}{{\sqrt{x}}}={{b}^{c}}\\\\y=\ {{b}^{c}}\sqrt{x} \end{array}$

10.           Show that

              $\begin{array}{l} \text{(a)}\ \ \displaystyle\frac{1}{4} \log _{10} 8+\frac{1}{4} \log _{10} 2=\log _{10} 2\\\\ \text{(b)}\ \ 4 \log _{10} 3-2 \log _{10} 3+1=\log _{10} 90 \end{array}$

Show/Hide Solution

$\begin{array}{l}\text{(a)}\;\;\ \ \ \displaystyle \frac{1}{4}{{\log }_{{10}}}8+\displaystyle \frac{1}{4}{{\log }_{{10}}}2\\\ \ \ \ =\displaystyle \frac{1}{4}\left( {{{{\log }}_{{10}}}8+{{{\log }}_{{10}}}2} \right)\\\ \ \ \ =\displaystyle \frac{1}{4}{{\log }_{{10}}}\left( {8\times 2} \right)\\\ \ \ \ =\displaystyle \frac{1}{4}{{\log }_{{10}}}16\\\ \ \ \ =\displaystyle \frac{1}{4}{{\log }_{{10}}}{{2}^{4}}\\\ \ \ \ =\displaystyle \frac{1}{4}\times 4{{\log }_{{10}}}2\\\ \ \ \ ={{\log }_{{10}}}2\\\\\text{(b)}\;\;\ \ 4{{\log }_{{10}}}3-2{{\log }_{{10}}}3+1\\\ \ \ \ =2{{\log }_{{10}}}3+{{\log }_{{10}}}10\\\ \ \ \ ={{\log }_{{10}}}{{3}^{2}}+{{\log }_{{10}}}10\\\ \ \ \ ={{\log }_{{10}}}\left( {{{3}^{2}}\times 10} \right)\\\ \ \ \ ={{\log }_{{10}}}90\end{array}$

11.           Show that

              $\begin{array}{l} \text{(a)}\ \ a^{2 \log _{a} 3}+b^{3 \log _{b} 2}=17\\\\ \text{(b)}\ \ 3 \log _{6} 1296=2 \log _{4} 4096 \end{array}$

Show/Hide Solution

$\begin{array}{l}\text{(a)}\;\;\ \ {{a}^{{2{{{\log }}_{a}}3}}}+{{b}^{{3{{{\log }}_{b}}2}}}\\\ \ \ \ ={{a}^{{{{{\log }}_{a}}{{3}^{2}}}}}+{{b}^{{{{{\log }}_{b}}{{2}^{3}}}}}\\\ \ \ \ ={{3}^{2}}+{{2}^{3}}\\\ \ \ \ =17\\\\\text{(b)}\;\ \ 3{{\log }_{6}}1296=3{{\log }_{6}}{{6}^{4}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3\times 4{{\log }_{6}}6\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =12\\\ \ \ \ \ \ 2{{\log }_{4}}4096=2{{\log }_{4}}{{4}^{6}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\times 6{{\log }_{4}}4\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =12\\\ \ \ \therefore \ \ 3{{\log }_{6}}1296=2{{\log }_{4}}4096\end{array}$

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

Show/Hide Solution

$\begin{array}{l}{{\log }_{{10}}}12=1.0792\\\\{{\log }_{{10}}}24=1.3802\\\\{{\log }_{{10}}}2={{\log }_{{10}}}\displaystyle \frac{{24}}{{12}}\\\ \ \ \ \ \ \ ={{\log }_{{10}}}24-{{\log }_{{10}}}12\\\ \ \ \ \ \ \ =1.3802-1.0792\\\ \ \ \ \ \ \ =0.3010\\\\{{\log }_{{10}}}6={{\log }_{{10}}}\displaystyle \frac{{12}}{2}\\\ \ \ \ \ \ \ ={{\log }_{{10}}}12-{{\log }_{{10}}}2\\\ \ \ \ \ \ \ =1.0792-0.3010\\\ \ \ \ \ \ \ =0.7782\end{array}$

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

Show/Hide Solution

$\begin{array}{l}{{\log }_{x}}a=5\\\\\therefore \ \ a={{x}^{5}}\\\\{{\log }_{x}}3a=9\\\\\therefore \ \ 3a={{x}^{9}}\\\\\therefore \ \ 3{{x}^{5}}={{x}^{9}}\\\\\therefore \ \ {{x}^{4}}=3\\\\\therefore \ \ x={{3}^{{\frac{1}{4}}}}=\sqrt[4]{3}\\\\\therefore \ \ a={{\left( {{{3}^{{\frac{1}{4}}}}} \right)}^{5}}={{3}^{{\frac{5}{4}}}}=3\sqrt[4]{3}\end{array}$

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

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

Show/Hide Solution

$\begin{array}{l}(\text{a})\ {{\log }_{{10}}}2=a\ (\text{given})\\\\\ \ \ {{\log }_{{10}}}8+{{\log }_{{10}}}25\\\\={{\log }_{{10}}}8+{{\log }_{{10}}}\frac{{100}}{4}\\\\={{\log }_{{10}}}8+{{\log }_{{10}}}100-{{\log }_{{10}}}4\\\\={{\log }_{{10}}}{{2}^{3}}+{{\log }_{{10}}}{{10}^{2}}-{{\log }_{{10}}}{{2}^{2}}\\\\=3{{\log }_{{10}}}2+2{{\log }_{{10}}}10-2{{\log }_{{10}}}2\\\\={{\log }_{{10}}}2+2\\\\=a+2\\\\(\text{b})\ \ \left. \begin{array}{l}a={{10}^{x}}\\b={{10}^{y}}\end{array} \right\}(\text{given})\\\\\ \ \ \ \ {{\log }_{{10}}}\left( {{{a}^{4}}{{b}^{3}}} \right)={{\log }_{{10}}}\left( {{{{\left( {{{{10}}^{x}}} \right)}}^{4}}{{{\left( {{{{10}}^{y}}} \right)}}^{3}}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{{10}}}\left( {\left( {{{{10}}^{{4x}}}} \right)\left( {{{{10}}^{3}}^{y}} \right)} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{{10}}}{{10}^{{4x+3y}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4x+3y\end{array}$

15.           $\text { (a) }$ If $\log _{2}(4 x-4)=2$, find the value of $\log _{4} x$.

   $\text { (b) }$ Prove that if $\displaystyle\frac{1}{2} \log _{3} M+3 \log _{3} N=1$ then $M N^{6}=9$.

Show/Hide Solution

$\begin{array}{l} (\text{a})\ {{\log }_{2}}(4x-4)=2\\\\ \ \ \ 4x-4={{2}^{2}}\\\\ \ \ \ 4x=8\\\\ \ \ \ x=2\\\\ \ \ \ x={{4}^{{\frac{1}{2}}}}\\\\ \ \ \ {{\log }_{4}}x=\displaystyle\frac{1}{2}\\\\\\ (\text{b})\ \ \displaystyle\frac{1}{2}{{\log }_{3}}M+3{{\log }_{3}}N=1\\\\ \ \ \ \ \ {{\log }_{3}}{{M}^{{\frac{1}{2}}}}+{{\log }_{3}}{{N}^{3}}=1\\\\ \ \ \ \ \ {{\log }_{3}}\left( {{{M}^{{\frac{1}{2}}}}{{N}^{3}}} \right)=1\\\\ \ \ \ \ \ {{M}^{{\frac{1}{2}}}}{{N}^{3}}=3\\\\ \ \ \ \ \ \text{Squaring both sides}\text{.}\\\\ \ \ \ \ \ M{{N}^{6}}=9 \end{array}$

စာဖတ်သူ၏ အမြင်ကို လေးစားစွာစောင့်မျှော်လျက်!