Question

If $1, log_{81} \left(3^x + 48\right)$ and $log_{9}\left(3^{x} - \frac{8}{3}\right) $ are in $A. P$., then $x$ is equal to

• A. $1$
• B. $2$
• C. $9$
• D. $3$

Answer 1

Answer: (B)

The three numbers $log_{9} 9, log_{9^{2}} \left(3^{x} +48\right)$ and $log_{9} \left(3^{x} - \frac{8}{3}\right)$ are in $A. P$.
i.e. $log_{9}9, \frac{1}{2} log_{9} \left(3^{x} + 48\right), log_{9} \left(3^{x}-\frac{8}{3} \right) $ are in $A. P$.
$ \Rightarrow 3^{x} + 48 = 9\left(3^{x} -\frac{8}{3}\right)$
$ \Rightarrow 8 \times 3^{x} =72 $
$ \Rightarrow 3^{x} = 9 $
$ \Rightarrow x = 2 $.