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)

$2$

Answer 2

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 $A. P$. $\Rightarrow 3^{x} + 48 = 9\left(3^{x} -\frac{8}{3}\right)$ $\Rightarrow 8 \times 3^{x} =72$ $\Rightarrow 3^{x} = 9$ $\Rightarrow x = 2$.