Question 1141069
Compounded ANNUALLY.


{{{1.036^1=1.036}}}
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{{{1.036^2=1.0733}}}
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{{{1.036^4=1.15196}}}
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{{{1.036^8=1.327}}}
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{{{1.036^11=1.476}}}
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{{{highlight(n=20)}}}


{{{1.036^20=2.029}}}




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{{{2=1.036^n}}}
{{{log((2))=n*log((1.036))}}}

{{{n=log((2))/log((1.036))}}}

{{{n=19.60}}}, close to 20 is better.