Answer :
Assuming [tex]\log=\ln[/tex] :
[tex]\log((5x-1)^4)=\frac{1}3\\ 4\log(5x-1)=\frac{1}3\\ 5x-1=e^{\frac{1}{12}}\\ \boxed{x=\dfrac{e^{\frac{1}{12}}+1}{5}}[/tex]
[tex]\log((5x-1)^4)=\frac{1}3\\ 4\log(5x-1)=\frac{1}3\\ 5x-1=e^{\frac{1}{12}}\\ \boxed{x=\dfrac{e^{\frac{1}{12}}+1}{5}}[/tex]
