Evaluate: $\underset{x→0}{\lim}\frac{\log(1-3x)}{5^x-1}$

$\frac{-3}{\log 5}$

$\underset{x→0}{\lim}\frac{ln(1-3x)}{5^x-1}=\underset{x→0}{\lim}\frac{(-3x)(-\frac{1}{3x})ln(1-3x)}{(\frac{5^x-1}{x}).x}=\underset{x→0}{\lim}\frac{-3\,ln(1-3x)^{-1/3x}}{\frac{5^x-1}{x}}$

$=\frac{-3}{ln\,5}[\underset{t→0}{lt}(1+t)^{1/t}=e\,and\,\underset{h→0}{lt}\frac{a^h-1}{h}=ln\,a]$