Logarithm — Part 1 MCQs 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 $$\eqalign{ & = {\log _{60}}3 + {\log _{60}}4 + {\log _{60}}5 \cr & = {\log _{60}}\left( {3 \times 4 \times 5} \right) \cr & = {\log _{60}}60 \cr & = 1 \cr} $$

$$\eqalign{ & \log \left( {{2^{64}}} \right) \cr & = 64 \times \log 2 \cr & = \left( {64 \times 0.30103} \right) \cr & = 19.26592 \cr} $$ Its characteristic is 19. Hence, then number of digits in 2 64 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 log 2 16 = n. Then, 2 n = 16 = 2 4 ⇒ n = 4 ∴ log 2 16 = 4

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$\eqalign{ & a = {\log _8}225 \cr & = {\log _{{2^3}}}\left( {{{15}^2}} \right) \cr & = \frac{2}{3}{\log _2}15 \cr & = \frac{{2b}}{3} \cr} $$