Logarithm - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{let }}{\log _{10000}}x = - \frac{1}{4} \cr & \Rightarrow x = {\left( {10000} \right)^{ - \frac{1}{4}}} \cr & = {\left( {{{10}^4}} \right)^{ - \frac{1}{4}}} \cr & = {10^{ - 1}} \cr & = \frac{1}{{10}} \cr} $$

$$\eqalign{ & \frac{{\log \sqrt 8 }}{{\log 8}} \cr & = \frac{{\log {{\left( 8 \right)}^{\frac{1}{2}}}}}{{\log 8}} \cr & = \frac{{\frac{1}{2}\log 8}}{{\log 8}} \cr & = \frac{1}{2} \cr} $$

$$\eqalign{ & \frac{{6{\text{ }}{{\log }_{10}}1000}}{{3{\text{ }}{{\log }_{10}}100}} \cr & = \frac{{6{\text{ }}{{\log }_{10}}{{10}^3}}}{{3{\text{ }}{{\log }_{10}}{{10}^2}}} \cr & = \frac{{6 \times 3{\text{ }}{{\log }_{10}}10}}{{3 \times 2{\text{ }}{{\log }_{10}}10}} \cr & = \frac{{18}}{6} \cr & = 3 \cr} $$

$$\eqalign{ & {{\log }_2}\left[ {{\text{ }}{{\log }_3}\left( {{\text{ }}{{\log }_2}x} \right)} \right] = 1 \cr & \Rightarrow {\log _3}\left( {{{\log }_2}x} \right) = {2^1} = 2 \cr & \Rightarrow {\log _2}x = {3^2} = 9 \cr & \Rightarrow x = {2^9} = 512 \cr} $$

$$\eqalign{ & \log \left( {\frac{9}{{14}}} \right) - \log \left( {\frac{{15}}{{16}}} \right) + \log \left( {\frac{{35}}{{24}}} \right) \cr & = \log \left( {\frac{9}{{14}} \div \frac{{15}}{{16}} \times \frac{{35}}{{24}}} \right) \cr & = \log \left( {\frac{9}{{14}} \times \frac{{16}}{{15}} \times \frac{{35}}{{24}}} \right) \cr & = \log 1 \cr & = 0 \cr} $$

$$\eqalign{ & 2{\log _{10}}5 + {\log _{10}}8 - \frac{1}{2}{\log _{10}}4 \cr & = {\log _{10}}\left( {{5^2}} \right) + {\log _{10}}8 - {\log _{10}}\left( {{4^{\frac{1}{2}}}} \right) \cr & = {\log _{10}}25 + {\log _{10}}8 - {\log _{10}}2 \cr & = {\log _{10}}\left( {\frac{{25 \times 8}}{2}} \right) \cr & = {\log _{10}}100 \cr & = 2 \cr} $$

$$\eqalign{ & {\text{lo}}{{\text{g}}_a}\left( {ab} \right) = x \cr & \Rightarrow \frac{{\log ab}}{{\log a}} = x \cr & \Rightarrow \frac{{\log a + \log b}}{{\log a}} = x \cr & \Rightarrow 1 + \frac{{\log b}}{{\log a}} = x \cr & \Rightarrow \frac{{\log b}}{{\log a}} = x - 1 \cr & \Rightarrow \frac{{\log a}}{{\log b}} = \frac{1}{{x - 1}} \cr & \Rightarrow 1 + \frac{{\log a}}{{\log b}} = 1 + \frac{1}{{x - 1}} \cr & \Rightarrow \frac{{\log b}}{{\log b}} + \frac{{\log a}}{{\log b}} = \frac{x}{{x - 1}} \cr & \Rightarrow \frac{{\log b + \log a}}{{\log b}} = \frac{x}{{x - 1}} \cr & \Rightarrow \frac{{{\text{log}}\left( {ab} \right)}}{{\log b}} = \frac{x}{{x - 1}} \cr & \Rightarrow {\text{lo}}{{\text{g}}_b}\left( {ab} \right) = \frac{x}{{x - 1}} \cr} $$

$$\eqalign{ & {\log _5}12 = {\log _5}\left( {3 \times 4} \right) \cr & = {\log _5}3 + {\log _5}4 \cr & = {\log _5}3 + 2{\log _5}2 \cr & = \frac{{{{\log }_{10}}3}}{{{{\log }_{10}}5}} + \frac{{2{{\log }_{10}}2}}{{{{\log }_{10}}5}} \cr} $$
$$ = \frac{{{{\log }_{10}}3}}{{{{\log }_{10}}10 - {{\log }_{10}}2}}$$   $$ + \frac{{2{{\log }_{10}}2}}{{{{\log }_{10}}10 - {{\log }_{10}}2}}$$
$$\eqalign{ & = \frac{b}{{1 - a}} + \frac{{2a}}{{1 - a}} \cr & = \frac{{2a + b}}{{1 - a}} \cr} $$

$$\eqalign{ & {\log _{10}}125 + {\log _{10}}8 = x \cr & \Rightarrow {\log _{10}}\left( {125 \times 8} \right) = x \cr & \Rightarrow x = {\log _{10}}\left( {1000} \right) \cr & \Rightarrow x = {\log _{10}}{\left( {10} \right)^3} \cr & \Rightarrow x = 3{\log _{10}}10 \cr & \Rightarrow x = 3 \cr} $$

$$\eqalign{ & {\log _5}\left( {{x^2} + x} \right) - {\log _5}\left( {x + 1} \right) = 2 \cr & \Rightarrow {\log _5}\left( {\frac{{{x^2} + x}}{{x + 1}}} \right) = 2\, \cr & \Rightarrow {\log _5}\left[ {\frac{{x\left( {x + 1} \right)}}{{x + 1}}} \right] = 2 \cr & \Rightarrow {\log _5}x = 2 \cr & \Rightarrow x = {5^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 25 \cr} $$