Question

If $1,lo{g}_{81}\left(3^x+48\right)$ and $lo{g}_9\left(3^x-\frac{8}{3}\right)$ are in $A.P$., then $x$ is equal to

• A. $1$
• B. $2$
• C. $9$
• D. $3$

Answer 1

Answer: (B)

The three numbers $lo{g}_{9}9,lo{g}_{9^2}\left(3^x+48\right)$ and $lo{g}_9\left(3^x-\frac{8}{3}\right)$ are in $A.P$.
i.e. $lo{g}_{9}9,\frac{1}{2}lo{g}_9\left(3^x+48\right),lo{g}_9\left(3^x-\frac{8}{3}\right)$ are in $A.P$.
$\Rightarrow 3^x+48=9\left(3^x-\frac{8}{3}\right)$
$\Rightarrow 8\times 3^x=72$
$\Rightarrow 3^x=9$
$\Rightarrow x=2$.