Question

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

Answer 1

$3f\left(x\right)-f\left(\frac{1}{x}\right)=\log e^{x^4};x>0$

Put $x=e^x\left\{f\right(e^x)=a,f(e^{-x})=6\}$

$3f\left(e^x\right)-f\left(e^{-x}\right)={\log }_e{e}^{4x}$

$3a-b=4x-\left(1\right)$

Put $x=e^{-x}$

$3f\left(e^{-x}\right)-f\left(e^x\right)={\log }_e{e}^{-4x}$

$3b-a=-4x-\left(2\right)$

Put $b=3a-4x$ from (1) in (2).

$3(3a-4x)-a=-4x$

$9a-12x-a=-4x$

$8a=8x$

$a=x$

$f\left(e^x\right)=x$