Question 1067732
{{{X=((Y/Z)^(1/n))-1}}}



Just reverse the order of everything done to n.


{{{X+1=(Y/Z)^(1/n)}}}
but from here, the inverse of exponentiation is logarithm.


{{{log((X+1))=log(((Y/Z)^(1/n)))}}}


{{{log((X+1))=(1/n)*log((Y/Z))}}}


{{{n*log((X+1))=log((Y/Z))}}}


{{{n=log((Y/Z))/log((X+1))}}}