Question

If $xϵR$ and $3^x+3^{-x}={\log }_{5}9$ then the value of $x$ equals

• A. $2$
• B. $3$
• C. $9$
• D. None of these

Answer 1

Answer: (D)

None of these

$3^x+3^{-x}={\log }_{5}9$

$\Rightarrow 3^x+\frac{1}{3^x}={\log }_{5}9$

$\Rightarrow 3^x+\frac{1}{3^x}=2.{\log }_{5}3=2.\frac{{\log }_{10}3}{{\log }_{10}5}$

$\Rightarrow 3^x+\frac{1}{3^x}=2.\frac{\left(0.4771\right)}{1-{\log }_{10}2}$

Take $3^x=t$

$\Rightarrow t+\frac{1}{t}=2\times \left\{\frac{0.4771}{0.6990}\right\}<2$

$\Rightarrow t+\frac{1}{t}<2\to$Not possible

Hence, no solution for x.