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If $$\log 3\log \left( {{3^x} - 2} \right)\,$$   and $$\log \left( {{3^x} + 4} \right)$$   are in arithmetic progression, then x is equal to

A. $$\frac{8}{3}$$

B. $$\log {3^8}$$

C. $$\log {2^3}$$

D. 8

Answer: Option C

Solution(By Examveda Team)

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

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