Logarithm - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & \frac{1}{2}\left( {\log x + \log y} \right) = \log \left( {\frac{{x + y}}{2}} \right) \cr & \Rightarrow \frac{1}{2}\log \left( {xy} \right) = \log \left( {\frac{{x + y}}{2}} \right) \cr & \Rightarrow \log {\left( {xy} \right)^{\frac{1}{2}}} = \log \left( {\frac{{x + y}}{2}} \right) \cr & \Rightarrow {\left( {xy} \right)^{\frac{1}{2}}} = \left( {\frac{{x + y}}{2}} \right) \cr & \Rightarrow xy = {\left( {\frac{{x + y}}{2}} \right)^2} \cr & \Rightarrow 4xy = {x^2} + {y^2} + 2xy \cr & \Rightarrow {x^2} + {y^2} - 2xy = 0 \cr & \Rightarrow {\left( {x - y} \right)^2} = 0 \cr & \Rightarrow x - y = 0 \cr & \Rightarrow x = y\, \cr} $$

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

$$\eqalign{ & {\text{Given}}\,\,\,{\text{Expression }} \cr & = {\text{ }}\log \left( {\frac{{{a^2}}}{{bc}} \times \frac{{{b^2}}}{{ac}} \times \frac{{{c^2}}}{{ab}}} \right) \cr & = \log 1 \cr & = 0 \cr} $$

$$\eqalign{ & {\text{Given}}\,\,\,{\text{Expression}} \cr & \left( {\frac{{\log a}}{{\log b}} \times \frac{{\log b}}{{\log c}} \times \frac{{\log c}}{{\log a}}} \right) \cr & = 1 \cr} $$

$$\eqalign{ & {\text{Given}}\,\,\,{\text{Expression}} \cr & = \frac{1}{{{{\log }_a}bc + {{\log }_a}a}} + \frac{1}{{{{\log }_b}ca + {{\log }_b}b}} + \frac{1}{{{{\log }_c}ab + {{\log }_c}c}} \cr & = \frac{1}{{{{\log }_a}\left( {abc} \right)}} + \frac{1}{{{{\log }_b}\left( {abc} \right)}} + \frac{1}{{{{\log }_c}\left( {abc} \right)}} \cr & = {\log _{abc}}a + {\log _{abc}}b + {\log _{abc}}c \cr & = {\log _{abc}}\left( {abc} \right) \cr & = 1 \cr} $$

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

$$\eqalign{ & \log x - 5\log 3 = - 2 \cr & \Rightarrow \log x - \log {3^5} = - 2 \cr & \Rightarrow \log \left( {\frac{x}{{{3^5}}}} \right) = - 2 \cr & \Rightarrow \frac{x}{{243}} = {10^{ - 2}} = \frac{1}{{100}} \cr & \Rightarrow x = \frac{{243}}{{100}} = 2.43 \cr} $$

$$\eqalign{ & a = {b^2} = {c^3} = {d^4} \cr & \Rightarrow b = {a^{\frac{1}{2}}},\,\,\,\,c = {a^{\frac{1}{3}}},\,\,\,\,d = {a^{\frac{1}{4}}} \cr & \therefore {\log _a}\left( {abcd} \right) \cr & = {\log _a}\left( {a \times {a^{\frac{1}{2}}} \times {a^{\frac{1}{3}}} \times {a^{\frac{1}{4}}}} \right) \cr & = {\log _a}{a^{\left( {1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}} \right)}} \cr & = \left( {1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}} \right){\log _a}a \cr & = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \cr} $$

$$\eqalign{ & \Rightarrow {\log _3}x + {\log _9}{x^2} + {\log _{27}}{x^3} = 9 \cr & \Rightarrow {\log _3}x + {\log _{{3^2}}}{x^2} + {\log _{{3^3}}}{x^3} = 9 \cr & \Rightarrow {\log _3}x + \frac{2}{2}{\log _3}x + \frac{3}{3}{\log _3}x = 9 \cr & \Rightarrow 3{\log _3}x = 9 \cr & \Rightarrow {\log _3}x = 3 \cr & \Rightarrow x = {3^3} = 27 \cr} $$

$$\eqalign{ & \,{\log _7}{\log _5}\left( {\sqrt {x + 5} + \sqrt x } \right) = 0 \cr & \Rightarrow {\log _5}\left( {\sqrt {x + 5} + \sqrt x } \right) = {7^0} = 1 \cr & \Rightarrow \sqrt {x + 5} + \sqrt x = {5^1} = 5 \cr & \Rightarrow {\left( {\sqrt {x + 5} + \sqrt x } \right)^2} = 25 \cr & \Rightarrow \left( {x + 5} \right) + x + 2\sqrt {x + 5} \sqrt x = 25 \cr & \Rightarrow 2x + 2\sqrt{ x}\,\, \sqrt {x + 5} = 20 \cr & \Rightarrow \sqrt {x}\,\, \sqrt {x + 5} = 10 - x \cr & \Rightarrow x\left( {x + 5} \right) = {\left( {10 - x} \right)^2} \cr & \Rightarrow {x^2} + 5x = 100 + {x^2} - 20x \cr & \Rightarrow 25x = 100 \cr & \Rightarrow x = 4 \cr} $$