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 $2,\frac{log4}{log\left(3^x-4\right)},\frac{log4}{log\left(3^x+\frac{7}{2}\right)}$ are in A.P.
$\Rightarrow \frac{1}{2},\frac{log\left(3^x-4\right)}{log4},\frac{log\left(3^x+\frac{7}{2}\right)}{log4}$ are in A.P.
$\Rightarrow \frac{1}{2}log4,log\left(3^x-4\right),log\left(3^x+\frac{7}{2}\right)$ are in A.P.
$\Rightarrow log2,log\left(3^x-4\right),log\left(3^x+\frac{7}{2}\right)$ 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.