Logarithms : Exercise (3.2) Solutions

1.           Write the following equations in logarithmic form.

              (a)   $3^{4}=81$

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

              (c)   $10^{-3}=0.001$

              (d)   $3^{-1}=\displaystyle\frac{1}{3}$

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

Show/Hide Solution

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

2.           Write the following equations in exponential form.

             (a)   $\log _{10} 3=0.4771$

             (b)   $\log _{6} 0.001=-3.855$

             (c)   $\log _{144} 12=\displaystyle\frac{1}{2}$

             (d)   $-5=\log _{3}\displaystyle \frac{1}{243}$

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

Show/Hide Solution

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

3.           Solve the following equations

             (a)   $\log _{7} 49=x$

             (b)   $\log _{x} 10=1$

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

             (d)   $x^{\log _{x} x}=5$

             (e)   $\log _{0.2} 5=x$

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$\begin{array}{l} \text{(a)}\ \ {{\log }_{7}}49=x\\ \ \ \ \ \ {{7}^{x}}=49\\ \ \ \ \ \ {{7}^{x}}={{7}^{2}}\\ \ \ \ \ \ x=2\\\\ \text{(b)}\ \ {{\log }_{x}}10=1\\ \ \ \ \ \ {{x}^{1}}=10\\ \ \ \ \ \ x=10\\\\ \text{(c)}\ \ {{\log }_{{\sqrt{3}}}}x=2\ \\ \ \ \ \ x={{\left( {\sqrt{3}} \right)}^{2}}\\ \ \ \ \ \ x=3\\\\ \text{(d)}\ \ {{x}^{{{{{\log }}_{x}}x}}}=5\\ \ \ \ \ \ x=5\\\\ \text{(e)}\ \ {{\log }_{{0.2}}}5=x\\ \ \ \ \ \ {{0.2}^{x}}=5\\ \ \ \ \ \ {{\left( {\displaystyle\frac{1}{5}} \right)}^{x}}=5\\ \ \ \ \ \ {{5}^{{-x}}}={{5}^{1}}\\\ \ \ \ \ x=-1\ \ \end{array}$

4.           Evaluate.

             (a)   $9^{\log _{9} 2}+3^{\log _{3} 8}$

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

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

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

             (e)   $10^{1-\log _{10} 3}$

Show/Hide Solution

$\begin{array}{l} \text{(a)}\ \ \ \ {{9}^{{{{{\log }}_{9}}2}}}+{{3}^{{{{{\log }}_{3}}8}}}\\ \ \ \ \ \ =2+8\\\ \ \ \ \ =10\\\\ \text{(b)}\ \ \ \ {{\log }_{4}}{{4}^{5}}\times {{\log }_{{10}}}{{10}^{2}}\\ \ \ \ \ \ =5\times 2\\\ \ \ \ \ =10\\\\ \text{(c)}\ \ \ \ \ {{7}^{{{{{\log }}_{7}}9}}}+{{\log }_{2}}\left( {\displaystyle\frac{1}{2}} \right)\\ \ \ \ \ \ ={{7}^{{{{{\log }}_{7}}9}}}+{{\log }_{2}}{{2}^{{-1}}}\\ \ \ \ \ \ =9-1\\\ \ \ \ \ =8\\\\ \text{(d)}\ \ \ \ \ {{\log }_{{\frac{1}{2}}}}\displaystyle\frac{1}{8}-4\cdot {{\log }_{{10}}}10\\ \ \ \ \ \ ={{\log }_{{\frac{1}{2}}}}{{\left( {\displaystyle\frac{1}{2}} \right)}^{3}}-4\cdot {{\log }_{{10}}}10\\ \ \ \ \ \ =3-4\cdot (1)\\\ \ \ \ \ =-1\\\\ \text{(e)}\ \ \ \ \ {{10}^{{1-{{{\log }}_{{10}}}3}}}\\ \ \ \ \ \ =\displaystyle\frac{{10}}{{{{{10}}^{{{{{\log }}_{{10}}}3}}}}}\\ \ \ \ \ \ =\displaystyle\frac{{10}}{3} \end{array}$

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

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

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

             (c)   $8+3^{x}=10,$ given that $\log _{3} 2=0.6309$

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$\begin{array}{l}\text{(a)}\ \ {{\log }_{3}}(2x-5)=2,\text{where}\ x>\displaystyle \frac{5}{2}\\\ \ \ \ \ 2x-5={{3}^{2}}\\\ \ \ \ \ 2x-5=9\\\ \ \ \ \ x=7\\\\\text{(b)}\ \ {{\log }_{{77}}}\left( {{{{\log }}_{7}}x} \right)=0,\ \ \text{where}\ x>0\\\ \ \ \ \ {{\log }_{7}}x={{77}^{0}}\\\ \ \ \ \ {{\log }_{7}}x=1\\\ \ \ \ \ x={{7}^{1}}=7\\\\\text{(c)}\ \ 8+{{3}^{x}}=10,\ \text{given that}\ {{\log }_{3}}2=0.6309\\\ \ \ \ \ {{3}^{x}}=2\\\ \ \ \ \ x={{\log }_{3}}2\\\ \ \ \ \ x=0.6309\end{array}$

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