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