Logarithm - Aptitude MCQ Questions and Solutions with Explanations

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

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

$$\eqalign{ & {\text{let }}\log x = - 3.153 \cr & {\text{then, }}\log x = - 3.153 \cr & = - 3 + \left( { - 0.153} \right) \cr & = \left( { - 3 - 1} \right) + \left( {1 - 0.153} \right) \cr & = - 4 + 0.847 = \overline 4 .847 \cr & {\text{Hence,}} \cr & {\text{characteristic = - 4, mantissa = 0}}{\text{.847 }} \cr} $$

$$\eqalign{ & \log {4^{50}} \cr & = 50\log 4 \cr & = 50\log {2^2} \cr & = \left( {50 \times 2} \right)\log 2 \cr & = 100 \times \log 2 \cr & = \left( {100 \times 0.30103} \right) \cr & = 30.103 \cr & \therefore {\text{characteristic}} = 30, \cr} $$
Hence, the number of digits in $${{\text{4}}^{50}} = 31$$

$$\eqalign{ & \log \left( {{4^9} \times {5^{17}}} \right) \cr & = \log \left( {{4^9}} \right) + \log \left( {{5^{17}}} \right) \cr & = \log \left( {{2^2}} \right)^9 + \log \left( {{5^{17}}} \right) \cr & = \log \left( {{2^{18}}} \right) + \log \left( {{5^{17}}} \right) \cr & = 18\log 2 + 17\log 5 \cr & = 18\log 2 + 17\left( {\log 10 - \log 2} \right) \cr & = 18\log 2 + 17\log 10 - 17\log 2 \cr & = \log 2 + 17\log 10 \cr & = 0.3010 + 17 \times 1 = 17.3010 \cr & \therefore {\text{Characteristic}} = 17 \cr & {\text{Hence, the number of digits in }}\left( {{4^9} \times {5^{17}}} \right) = 18 \cr} $$

In arithmetic progression common ratio are equal to
$$\log \left( {{3^x} - 2} \right) - \log 3 = $$     $$\log \left( {{3^x} + 4} \right) - $$   $$\log \left( {{3^x} - 2} \right)$$
$$\frac{{\log \left( {{3^x} - 2} \right)}}{{\log 3}} = \frac{{\log \left( {{3^x} + 4} \right)}}{{\log \left( {{3^x} - 2} \right)}}$$     $$\left( {\therefore \log a - \log b = \log \frac{a}{b}} \right)$$
$$\eqalign{ & \frac{{\log {3^x}}}{{\log 2\log 3}} = \frac{{x\log 3\log 4\log 2}}{{x\log 3}} \cr & \frac{{x\log 3}}{{\log 2\log 3}} = \frac{{x\log 3\log 4\log 2}}{{x\log 3}} \cr & \frac{x}{{\log 2}} = \log 4\log 2 \cr & x = \log 4\log \,2\log 2 \cr & x = \log 8 \cr & x = \log {2^3} \cr} $$

$$\eqalign{ & {\text{Given}}, \cr & {\log _{10}}a = p,\,{\log _{10}}b = q \cr & {\log _{10}}\left( {{a^p}{b^q}} \right) = {\log _{10}}{a^p} + {\log _{10}}{b^q} \cr & = p{\log _{10}}a + q{\log _{10}}b \cr & = {p^2} + {q^2} \cr} $$