Logarithm - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option C

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

Answer: Option A

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