Question

The solution of equation
$9^x-2^{x+\frac{1}{2}}=2^{x+\frac{3}{2}}-3^{2x-1}$ is

• A. ${\log }_e\left(\frac{9}{\sqrt{8}}\right)$
• B. ${\log }_9\left(\frac{9}{\sqrt{8}}\right)$
• C. ${\log }_{(9/2)}\left(\frac{9}{\sqrt{8}}\right)$
• D. None of these

Answer 1

Answer: (C)

Given, equation is
$9^x-2^{x+\frac{1}{2}}=2^{x+\frac{3}{2}}-3^{2x-1}$
$\Rightarrow$ $9^x-2^x.\sqrt{2}=2^x\sqrt{8}-9^x{.3}^{-1}$
$\Rightarrow$ $9^x+\frac{9^x}{3}={2.2}^x\sqrt{2}+2^x\sqrt{2}$
$\Rightarrow$ $\frac{4}{3}{9}^x=3\sqrt{2}{.2}^x$
$\Rightarrow$ ${\left(\frac{9}{2}\right)}^x=\frac{9\sqrt{2}}{4}$
$\Rightarrow$ ${\left(\frac{9}{2}\right)}^x=\frac{9}{2\sqrt{2}}=\frac{9}{\sqrt{8}}$
$\Rightarrow$ $x={\log }_{(9/2)}\left(\frac{9}{\sqrt{8}}\right)$