Quantitative Aptitude
LOGARITHM MCQs
Logarithms
Total Questions : 289
| Page 26 of 29 pages
Answer: Option C. -> $$\frac{{\log a}}{{\log b}} = \frac{y}{x}$$
$$\eqalign{
& {a^x} = {b^y} \cr
& \Rightarrow \log {a^x} = \log {b^y} \cr
& \Rightarrow x\log a = y\log b \cr
& \Rightarrow {{\log a} \over {\log b}} = {y \over x} \cr} $$
$$\eqalign{
& {a^x} = {b^y} \cr
& \Rightarrow \log {a^x} = \log {b^y} \cr
& \Rightarrow x\log a = y\log b \cr
& \Rightarrow {{\log a} \over {\log b}} = {y \over x} \cr} $$
Answer: Option B. -> 4
Let log2 16 = n.
Then, 2n = 16 = 24
⇒ n = 4
∴ log2 16 = 4
Let log2 16 = n.
Then, 2n = 16 = 24
⇒ n = 4
∴ log2 16 = 4
Answer: Option C. -> 21000
$$\eqalign{
& {\log _2}x = 10\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,x = {2^{10}} \cr
& \therefore {\log _x}y = 100 \cr
& \Rightarrow y = {x^{100}} \cr
& \Rightarrow y = {\left( {{2^{10}}} \right)^{100}}\,\,\,\left[ {{\text{put}}\,{\text{value}}\,{\text{of}}\,x} \right] \cr
& \Rightarrow y = {2^{1000}} \cr} $$
$$\eqalign{
& {\log _2}x = 10\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,x = {2^{10}} \cr
& \therefore {\log _x}y = 100 \cr
& \Rightarrow y = {x^{100}} \cr
& \Rightarrow y = {\left( {{2^{10}}} \right)^{100}}\,\,\,\left[ {{\text{put}}\,{\text{value}}\,{\text{of}}\,x} \right] \cr
& \Rightarrow y = {2^{1000}} \cr} $$
Answer: Option D. -> $$\frac{{256}}{{81}}$$
$$\eqalign{
& {\log _x}\left( {{9 \over {16}}} \right) = - {1 \over 2} \cr
& \Rightarrow {x^{ - {1 \over 2}}} = {9 \over {16}} \cr
& \Rightarrow {1 \over {\sqrt x }} = {9 \over {16}} \cr
& \Rightarrow \sqrt x = {{16} \over 9} \cr
& \Rightarrow x = {\left( {{{16} \over 9}} \right)^2} \cr
& \Rightarrow x = {{256} \over {81}} \cr} $$
$$\eqalign{
& {\log _x}\left( {{9 \over {16}}} \right) = - {1 \over 2} \cr
& \Rightarrow {x^{ - {1 \over 2}}} = {9 \over {16}} \cr
& \Rightarrow {1 \over {\sqrt x }} = {9 \over {16}} \cr
& \Rightarrow \sqrt x = {{16} \over 9} \cr
& \Rightarrow x = {\left( {{{16} \over 9}} \right)^2} \cr
& \Rightarrow x = {{256} \over {81}} \cr} $$
Answer: Option A. -> $$\frac{1}{{10}}$$
$$\eqalign{
& {\text{let }}{\log _{10000}}x = - \frac{1}{4} \cr
& \Rightarrow x = {\left( {10000} \right)^{ - \frac{1}{4}}} \cr
& = {\left( {{{10}^4}} \right)^{ - \frac{1}{4}}} \cr
& = {10^{ - 1}} \cr
& = \frac{1}{{10}} \cr} $$
$$\eqalign{
& {\text{let }}{\log _{10000}}x = - \frac{1}{4} \cr
& \Rightarrow x = {\left( {10000} \right)^{ - \frac{1}{4}}} \cr
& = {\left( {{{10}^4}} \right)^{ - \frac{1}{4}}} \cr
& = {10^{ - 1}} \cr
& = \frac{1}{{10}} \cr} $$
Answer: Option D. -> 512
$$\eqalign{
& {{\log }_2}\left[ {{\text{ }}{{\log }_3}\left( {{\text{ }}{{\log }_2}x} \right)} \right] = 1 \cr
& \Rightarrow {\log _3}\left( {{{\log }_2}x} \right) = {2^1} = 2 \cr
& \Rightarrow {\log _2}x = {3^2} = 9 \cr
& \Rightarrow x = {2^9} = 512 \cr} $$
$$\eqalign{
& {{\log }_2}\left[ {{\text{ }}{{\log }_3}\left( {{\text{ }}{{\log }_2}x} \right)} \right] = 1 \cr
& \Rightarrow {\log _3}\left( {{{\log }_2}x} \right) = {2^1} = 2 \cr
& \Rightarrow {\log _2}x = {3^2} = 9 \cr
& \Rightarrow x = {2^9} = 512 \cr} $$
Answer: Option A. -> 0
$$\eqalign{
& \log \left( {\frac{9}{{14}}} \right) - \log \left( {\frac{{15}}{{16}}} \right) + \log \left( {\frac{{35}}{{24}}} \right) \cr
& = \log \left( {\frac{9}{{14}} \div \frac{{15}}{{16}} \times \frac{{35}}{{24}}} \right) \cr
& = \log \left( {\frac{9}{{14}} \times \frac{{16}}{{15}} \times \frac{{35}}{{24}}} \right) \cr
& = \log 1 \cr
& = 0 \cr} $$
$$\eqalign{
& \log \left( {\frac{9}{{14}}} \right) - \log \left( {\frac{{15}}{{16}}} \right) + \log \left( {\frac{{35}}{{24}}} \right) \cr
& = \log \left( {\frac{9}{{14}} \div \frac{{15}}{{16}} \times \frac{{35}}{{24}}} \right) \cr
& = \log \left( {\frac{9}{{14}} \times \frac{{16}}{{15}} \times \frac{{35}}{{24}}} \right) \cr
& = \log 1 \cr
& = 0 \cr} $$
Answer: Option D. -> 3
$$\eqalign{
& \frac{{6{\text{ }}{{\log }_{10}}1000}}{{3{\text{ }}{{\log }_{10}}100}} \cr
& = \frac{{6{\text{ }}{{\log }_{10}}{{10}^3}}}{{3{\text{ }}{{\log }_{10}}{{10}^2}}} \cr
& = \frac{{6 \times 3{\text{ }}{{\log }_{10}}10}}{{3 \times 2{\text{ }}{{\log }_{10}}10}} \cr
& = \frac{{18}}{6} \cr
& = 3 \cr} $$
$$\eqalign{
& \frac{{6{\text{ }}{{\log }_{10}}1000}}{{3{\text{ }}{{\log }_{10}}100}} \cr
& = \frac{{6{\text{ }}{{\log }_{10}}{{10}^3}}}{{3{\text{ }}{{\log }_{10}}{{10}^2}}} \cr
& = \frac{{6 \times 3{\text{ }}{{\log }_{10}}10}}{{3 \times 2{\text{ }}{{\log }_{10}}10}} \cr
& = \frac{{18}}{6} \cr
& = 3 \cr} $$
Answer: Option C. -> $$\frac{1}{2}$$
$$\eqalign{
& \frac{{\log \sqrt 8 }}{{\log 8}} \cr
& = \frac{{\log {{\left( 8 \right)}^{\frac{1}{2}}}}}{{\log 8}} \cr
& = \frac{{\frac{1}{2}\log 8}}{{\log 8}} \cr
& = \frac{1}{2} \cr} $$
$$\eqalign{
& \frac{{\log \sqrt 8 }}{{\log 8}} \cr
& = \frac{{\log {{\left( 8 \right)}^{\frac{1}{2}}}}}{{\log 8}} \cr
& = \frac{{\frac{1}{2}\log 8}}{{\log 8}} \cr
& = \frac{1}{2} \cr} $$
Answer: Option A. -> 2
$$\eqalign{
& 2{\log _{10}}5 + {\log _{10}}8 - \frac{1}{2}{\log _{10}}4 \cr
& = {\log _{10}}\left( {{5^2}} \right) + {\log _{10}}8 - {\log _{10}}\left( {{4^{\frac{1}{2}}}} \right) \cr
& = {\log _{10}}25 + {\log _{10}}8 - {\log _{10}}2 \cr
& = {\log _{10}}\left( {\frac{{25 \times 8}}{2}} \right) \cr
& = {\log _{10}}100 \cr
& = 2 \cr} $$
$$\eqalign{
& 2{\log _{10}}5 + {\log _{10}}8 - \frac{1}{2}{\log _{10}}4 \cr
& = {\log _{10}}\left( {{5^2}} \right) + {\log _{10}}8 - {\log _{10}}\left( {{4^{\frac{1}{2}}}} \right) \cr
& = {\log _{10}}25 + {\log _{10}}8 - {\log _{10}}2 \cr
& = {\log _{10}}\left( {\frac{{25 \times 8}}{2}} \right) \cr
& = {\log _{10}}100 \cr
& = 2 \cr} $$
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