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When solving the equation

$$150 = 160 - 40 e^{-t/20}$$

I come to a solution that seems natural to me as follows:

\begin{align*} .25 &= e^{-t/20}\\ \ln(.25) &= -t/20\\ t &= -20 \ln(.25)\\ &= 27.7259. \end{align*}

However, the textbook gives the same result as

$$t = 40 \ln(2) = 27.7259.$$

Can someone please explain to me how to derive the result from the book?

Regards, Danny.

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$$-20 \ln (.25) = -20 \ln (\frac 14) = 20 \ln (4) = 40 \ln 2$$

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We have $0.25=\frac 14=2^{-2}$, so $\log (0.25) = -2 \log (2)$

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  • $\begingroup$ yes, of course, $ln(1/x) = -ln(x)$ and $ln(a^b) = b ln(a)$ thank you both !!! $\endgroup$ – velsd Sep 4 at 11:09

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