$\begin{aligned} If \log 3 \log (3^{x} - 2) and \log (3^{x} + 4) \\ are in arithmetic progression, \\ then x is equal to - \end{aligned}$ $\eqalign{ & {\text{If}}\,\log 3\log \left( {{3^x} - 2} \right)\,{\text{and}}\,\log \left( {{3^x} + 4} \right)\, \cr & {\text{are}}\,{\text{in}}\,{\text{arithmetic}}\,\,{\text{progression}}\,{\text{,}} \cr & {\text{then}}\,x\,{\text{is}}\,{\text{equal}}\,{\text{to}} - \cr}$

1⁸/₃
2log 3⁸
3log 2³
48

Answer:- 1

Explanation:-

Solution:

\begin{aligned} In arithmetic progression common ratio are equal to \\ \log (3^{x} - 2) - \log 3 \\ = \log (3^{x} + 4) - \log (3^{x} - 2) \\ = \frac{\log (3^{x} - 2)}{\log 3} \\ = \frac{\log (3^{x} + 4)}{\log (3^{x} - 2)} (∴ \log a - \log b = \log \frac{a}{b}) \\ = \frac{\log 3^{x}}{\log 2 \log 3} \\ = \frac{x \log 3 \log 4 \log 2}{x \log 3} \\ = \frac{x \log 3}{\log 2 \log 3} \\ = \frac{x \log 3 \log 4 \log 2}{x \log 2} \\ = \frac{x}{\log 2} \\ = \log 4 \log 2 \\ \Rightarrow x = \log 4 \log 2 \log 2 \\ \Rightarrow x = \log 8 \\ \Rightarrow x = \log 2^{3} \end{aligned}

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