Logarithm - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{A = Since lo}}{{\text{g}}_a}a = 1,{\text{ so lo}}{{\text{g}}_{10}}10 = 1 \cr & \cr & B = \,{\text{log}}\left( {2 + 3} \right) = 5\, \cr & \,\,\,{\text{and log}}\left( {2 \times 3} \right) = {\text{log}}6 \cr & = {\text{log}}2 + {\text{log}}3 \cr & \therefore \,{\text{log}}\left( {2 + 3} \right) \ne {\text{log}}\left( {2 \times 3} \right). \cr & \cr & C = \,{\text{ Since lo}}{{\text{g}}_a}1 = 0,so{\text{ lo}}{{\text{g}}_{10}}1 = 0. \cr & \cr & {\text{D = log}}\left( {1 + 2 + 3} \right) = {\text{ log}}6 \cr & = {\text{ log}}\left( {1 \times 2 \times 3} \right) \cr & = {\text{ log}}1 + {\text{ log}}2 + {\text{ log}}3. \cr & {\text{So (B) is incorrect}} \cr} $$

$$\eqalign{ & {\log _5}512 \cr & = {{\log 512} \over {\log 5}} \cr & = {{\log {2^9}} \over {\log \left( {{{10} \over 2}} \right)}} \cr & = {{9\log 2} \over {\log 10 - \log 2}} \cr & = {{ {9 \times 0.3010} } \over {1 - 0.3010}} \cr & = {{2.709} \over {0.699}} \cr & = {{2709} \over {699}} \cr & = 3.876 \cr} $$

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

$$\eqalign{ & \log 27 = 1.431 \cr & \Rightarrow \log \left( {{3^3}} \right) = 1.431 \cr & \Rightarrow 3\,\log \,3 = 1.431 \cr & \Rightarrow \log \,3 = 0.477 \cr & \therefore \log \,9 = \log \left( {{3^2}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\,\log \,3 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {2 \times 0.477} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.954 \cr} $$

$$\eqalign{ & \log {a \over b} + \log {b \over a} = \log \left( {a + b} \right) \cr & \Rightarrow \log \left( {a + b} \right) = \log \left( {{a \over b} \times {b \over a}} \right) = \log 1 \cr & So,a + b = 1 \cr} $$

$$\eqalign{ & {\log _{10}}\left( {{1 \over {70}}} \right) \cr & = {\log _{10}}1 - {\log _{10}}70 \cr & = - {\log _{10}}\left( {7 \times 10} \right) \cr & = - \left( {{{\log }_{10}}7 + {{\log }_{10}}10} \right) \cr & = - \left( {a + 1} \right) \cr} $$

$$\eqalign{ & {\log _2}10 \cr & = \frac{1}{{{{\log }_{10}}2}} \cr & = \frac{1}{{0.3010}} \cr & = \frac{{10000}}{{3010}} \cr & = \frac{{1000}}{{301}} \cr} $$

$$\eqalign{ & {\log _{10}}80 \cr & = {\log _{10}}\left( {8 \times 10} \right) \cr & = {\log _{10}}8 + {\log _{10}}10 \cr & = {\log _{10}}\left( {{2^3}} \right) + 1 \cr & = 3{\log _{10}}2 + 1 \cr & = \left( {3 \times 0.3010} \right) + 1 \cr & = 1.9030 \cr} $$

$$\eqalign{ & \Rightarrow {\log _{10}}5 + {\log _{10}}\left( {5x + 1} \right) = {\log _{10}}\left( {x + 5} \right) + 1 \cr & \Rightarrow {\log _{10}}\left[ {5\left( {5x + 1} \right)} \right] = {\log _{10}}\left[ {10\left( {x + 5} \right)} \right] \cr & \Rightarrow 5\left( {5x + 1} \right) = 10\left( {x + 5} \right) \cr & \Rightarrow 5x + 1 = 2x + 10 \cr & \Rightarrow 3x = 9 \cr & \Rightarrow x = 3 \cr} $$

Given expression
$$\eqalign{ & = {\log _{60}}3 + {\log _{60}}4 + {\log _{60}}5 \cr & = {\log _{60}}\left( {3 \times 4 \times 5} \right) \cr & = {\log _{60}}60 \cr & = 1 \cr} $$