Question

If $3f\left(x\right)-f\left(\frac{1}{x}\right)={\log }_e{x}^4$ for $x>0$, then $f\left(e^x\right)$ is

• A. $2x$
• B. $x^2$
• C. $x$
• D. None of these

Answer 1

Answer: (C)

$x$

We have, $3f\left(x\right)-f\left(\frac{1}{x}\right)={\log }_e{x}^4$
$\Rightarrow 3f\left(x\right)-f\left(\frac{1}{x}\right)=4{\log }_{e}x....\left(i\right)$
$\Rightarrow 3f\left(\frac{1}{x}\right)-f\left(x\right)=4{\log }_e\left(\frac{1}{x}\right)$ (Replacing $x$ by $\frac{1}{x})$
$\Rightarrow 3f\left(\frac{1}{x}\right)-f\left(x\right)=-4{\log }_{e}x....\left(ii\right)$
Solving Eqs. (i) and (ii), we get
$8f\left(x\right)=8{\log }_{e}x\Rightarrow f\left(x\right)={\log }_{e}x$
$\Rightarrow f\left(e^x\right)=x$.