add 1 to both sides
\(\sqrt[5]{32^x-1}=1\)
raise each side to the 5th power
\(32^x-1=1\)
add 1 to both sides
\(32^x=2\)
we can try to match bases
because if, xᵃ=xᵇ where x=x, then a=b
we know that 32=2⁵
and \((x^m)^n=x^{mn}\) so
\(32^x=2\) is the same as
\((2^5)^x=2\) or \(2^{5x}=2^1\)
therefor
5x=1
divide both sides by 5
x=1/5
