Question 902938
Comfort with logarithms needed.


{{{A=50(1+r/12)^12}}}

{{{log((A))=log((50))+log(((1+r/12)^12))}}}

{{{log((A))=log((50))+12*log((1+r/12))}}}

{{{12*log((1+r/12))=log((A/50))}}}

{{{log((1+r/12))=(1/12)log((A/50))}}}


Decide what base, b, you want.
(This is not rendering neatly, so it might fail to appear meaningful/readable):


{{{b^((1/12)log((A/50)))=r/12+1}}}

{{{r/12=-1+b^((1/12)log((A/50)))}}}

{{{highlight(r=-12+12*b^((1/12)log((A/50))))}}}