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$$\log \left( {{3^x} - 2} \right) - \log 3 = $$ $$\log \left( {{3^x} + 4} \right) - $$ $$\log \left( {{3^x} - 2} \right)$$
$$\frac{{\log \left( {{3^x} - 2} \right)}}{{\log 3}} = \frac{{\log \left( {{3^x} + 4} \right)}}{{\log \left( {{3^x} - 2} \right)}}$$ $$\left( {\therefore \log a - \log b = \log \frac{a}{b}} \right)$$
$$\eqalign{ & \frac{{\log {3^x}}}{{\log 2\log 3}} = \frac{{x\log 3\log 4\log 2}}{{x\log 3}} \cr & \frac{{x\log 3}}{{\log 2\log 3}} = \frac{{x\log 3\log 4\log 2}}{{x\log 3}} \cr & \frac{x}{{\log 2}} = \log 4\log 2 \cr & x = \log 4\log \,2\log 2 \cr & x = \log 8 \cr & x = \log {2^3} \cr} $$
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Related Questions on Logarithm
Which of the following statements is not correct?
A. log10 10 = 1
B. log (2 + 3) = log (2 x 3)
C. log10 1 = 0
D. log (1 + 2 + 3) = log 1 + log 2 + log 3
$${{\log \sqrt 8 } \over {\log 8}}$$ is equal to:
A. $$\frac{1}{6}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$\frac{1}{8}$$
help me to understand in the 1st and 2nd lines.
how?