<PRE><B>Question 429743</B>
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A &lt;i&gt;&lt;b&gt;very&lt;/b&gt;&lt;/i&gt; long time.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ Pe^{rt}]


Substitute values


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5000e^{0.04t}\ =\ 30000]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{0.04t}\ =\ 6]


Take the natural log of both sides


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(e^{0.04t}\right)\ =\ \ln(6)]


Use the laws of logs and the fact that *[tex \Large \ln(e)\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.04t\ =\ \ln(6)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{\ln(6)}{0.04}\ \approx\ 44.8] years






John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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