Question 28474
{{{1.0201=(1+r/12)^12}}} two ways to solve this, the easiest being to take the twelth root of both sides with your calculator to get:
{{{root(12, 1.0201)=root(12, (1+r/12)^12)}}}
{{{root(12, 1.0201)=(1+r/12)}}}
{{{12(root(12, 1.0201)-1)=r}}}
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You can also do it with natural logs, but in this case you end up with a bunch of messy fractions and e^(ln(1.02...)/12), etc.... but at any rate, you should get:
0.019917
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{{{(1+.019917/12)^12=(1.00165975)^12=1.020100}}}