Question 964590
Yes, the first one is {{{A[1]=Pe^(rt)}}}
The second is {{{A[2]=P(1+i)^t}}}
Set them equal to each other,
{{{2000e^(0.07t)=16000(1.04)^t}}}
*[illustration ex1.JPG].
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Looks like it takes quite a while,
{{{t=67.6}}} years