Question

If the reciprocals of $2, \, \left(log\right)_{\left(3^{x} - 4\right)} 4$ and $\left(log\right)_{\left(3^{x} + \frac{7}{2}\right)} 4$ are in arithmetic progression, then $x$ is equal to

• A. $1$
• B. $2$
• C. $4$
• D. $0$

Answer 1

Answer: (B)

Reciprocals of $2,\frac{log 4}{log ⁡ \left(3^{x} - 4\right)},\frac{log ⁡ 4}{log ⁡ \left(3^{x} + \frac{7}{2}\right)}$ are in A.P.
$\Rightarrow \frac{1}{2},\frac{log \left(3^{x} - 4\right)}{log ⁡ 4},\frac{log ⁡ \left(3^{x} + \frac{7}{2}\right)}{log ⁡ 4}$ are in A.P.
$\Rightarrow \frac{1}{2}log 4,log ⁡ \left(3^{x} - 4 \right),log ⁡ \left(3^{x} + \frac{7}{2}\right)$ are in A.P.
$\Rightarrow log 2,log ⁡ \left(3^{x} - 4\right),log ⁡ \left(3^{x} + \frac{7}{2}\right)$ are in A.P.
$\Rightarrow 2,3^{x}-4,3^{x}+\frac{7}{2}$ are in G.P.
$\Rightarrow \left(3^{x} - 4\right)^{2}=2\left(3^{x} + \frac{7}{2}\right)$
Let $3^{x}=y$
$\Rightarrow \left(y - 4\right)^{2}=2\left(y + \frac{7}{2}\right)$
$\Rightarrow y^{2}+16-8y=2y+7$
$\Rightarrow y^{2}-10y+9=0$
$\Rightarrow \left(y - 1\right)\left(y - 9\right)=0\Rightarrow y=1,9$
$\Rightarrow 3^{x}=1,9\Rightarrow x=0,2$
But, for $x=0,3^{x}-4=-3 < 0$
So, $x=2$ will only form an A.P.