Logarithm - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & \log \left( {{2^{64}}} \right) \cr & = 64 \times \log 2 \cr & = \left( {64 \times 0.30103} \right) \cr & = 19.26592 \cr} $$
Its characteristic is 19.
Hence, then number of digits in 2⁶⁴ is 20.

$$\eqalign{ & {\log _x}\left( {{9 \over {16}}} \right) = - {1 \over 2} \cr & \Rightarrow {x^{ - {1 \over 2}}} = {9 \over {16}} \cr & \Rightarrow {1 \over {\sqrt x }} = {9 \over {16}} \cr & \Rightarrow \sqrt x = {{16} \over 9} \cr & \Rightarrow x = {\left( {{{16} \over 9}} \right)^2} \cr & \Rightarrow x = {{256} \over {81}} \cr} $$

$$\eqalign{ & {a^x} = {b^y} \cr & \Rightarrow \log {a^x} = \log {b^y} \cr & \Rightarrow x\log a = y\log b \cr & \Rightarrow {{\log a} \over {\log b}} = {y \over x} \cr} $$

$$\eqalign{ & {\log _2}x = 10\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,x = {2^{10}} \cr & \therefore {\log _x}y = 100 \cr & \Rightarrow y = {x^{100}} \cr & \Rightarrow y = {\left( {{2^{10}}} \right)^{100}}\,\,\,\left[ {{\text{put}}\,{\text{value}}\,{\text{of}}\,x} \right] \cr & \Rightarrow y = {2^{1000}} \cr} $$

Let log₂ 16 = n.
Then, 2ⁿ = 16 = 2⁴ ⇒ n = 4
∴ log₂ 16 = 4

$$\eqalign{ & {\text{lo}}{{\text{g}}_5}\frac{{\left( {125} \right)\left( {625} \right)}}{{25}} \cr & = {\text{lo}}{{\text{g}}_5}\left( {\frac{{{5^3} \times {5^4}}}{{{5^2}}}} \right) \cr & = {\text{lo}}{{\text{g}}_5}{5^5} \cr & {\text{ = 5lo}}{{\text{g}}_5}5 \cr & = 5 \cr} $$

$$\eqalign{ & {\text{lo}}{{\text{g}}_{3\sqrt 2 }}\left( {\frac{1}{{18}}} \right)\, \cr & = {\text{lo}}{{\text{g}}_{3\sqrt 2 }}\left( {\frac{1}{{{{\left( {3\sqrt 2 } \right)}^2}}}} \right)\, \cr & = {\text{lo}}{{\text{g}}_{3\sqrt 2 }}{\text{ }}{\left( {3\sqrt 2 } \right)^{ - 2}} \cr & = \left( { - 2} \right){\text{lo}}{{\text{g}}_{3\sqrt 2 }}3\sqrt 2 \cr & = - 2 \cr} $$

$$\eqalign{ & {\text{lo}}{{\text{g}}_{10}}\left( {0.0001} \right)\, \cr & = {\text{lo}}{{\text{g}}_{10}}\left( {\frac{1}{{10000}}} \right) \cr & \, = {\text{lo}}{{\text{g}}_{10}}\left( {\frac{1}{{{{10}^4}}}} \right) \cr & \, = {\text{lo}}{{\text{g}}_{10}}{10^{ - 4}} \cr & \, = - 4\,{\text{lo}}{{\text{g}}_{10}}10 \cr & = - 4{\text{ }} \cr} $$

$$\eqalign{ & {\left[ {{\text{lo}}{{\text{g}}_{10}}\left( {{\text{5lo}}{{\text{g}}_{10}}100} \right)\,} \right]^2} \cr & = {\left[ {{\text{lo}}{{\text{g}}_{10}}\left\{ {{\text{5lo}}{{\text{g}}_{10}}{{\left( {10} \right)}^2}} \right\}\,} \right]^2} \cr & = \,{\left[ {{\text{lo}}{{\text{g}}_{10}}\left( {5 \times 2} \right)} \right]^2} \cr & = {\left( {{\text{lo}}{{\text{g}}_{10}}10} \right)^2} \cr & = 1 \cr} $$

$$\eqalign{ & {\text{lo}}{{\text{g}}_8}{\text{p}} = 25\,\,{\text{and lo}}{{\text{g}}_2}{\text{q}} = 5\, \cr & \Rightarrow {\text{p = }}{{\text{8}}^{25}}\,{\text{and q}} = {2^5} \cr & \Rightarrow {\text{p = }}{\left( {{2^3}} \right)^{25}}{\text{ and q = }}{2^5} \cr & \Rightarrow {\text{p = }}{{\text{2}}^{75}}{\text{ and q = }}{2^5} \cr & \Rightarrow {\text{p = }}{\left( {{2^5}} \right)^{15}}{\text{ and q = }}{2^5} \cr & \Rightarrow {\text{ p = }}{{\text{q}}^{15}} \cr} $$