Question 145800
I'm presuming you mean: {{{S= C^(1-r)}}}


Take the log of both sides (doesn't matter what base)


{{{log(S)=log(C^(1-r))}}}


Since we know that {{{log(b,x^n)=n*log(b,x)}}}, we can say:


{{{log(S)=(1-r)log(C)}}}


Divide both sides by {{{log(C)}}}


{{{log(S)/log(C)=1-r}}}


Add {{{-1}}} to both sides


{{{log(S)/log(C)-1=-r}}}


Multiply both sides by {{{-1}}}


{{{r=1-log(S)/log(C)}}}