$\:a^x=e^{\ln \left(a^x\right)}=e^{x\cdot \ln \left(a\right)}$
= $\left(\frac{3}{4}\right)^x=e^{x\ln \left(\frac{3}{4}\right)}$
$=\lim _{x\to \infty \:}\left(e^{x\ln \left(\frac{3}{4}\right)}\right)$
$=\lim _{x\to \infty \:}\left(e^{x\ln \left(\frac{4}{3}\right)^{-1}}\right)$
$=\lim _{x\to \infty \:}\left(e^{-x\ln \left(\frac{4}{3}\right)}\right)$
= $e^{-\infty }$
Take x towards $-\infty$ ,$e^{-\infty }$ approaches zero .@your line and also did using log, but not getting equal to 0 . No chance that you will get $\left ( \frac{3}{4} \right )^{x} =0$. Hence proved Thank you !!