# $e$ and $\ln$ : how to derive two equivalent equations

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.

$$-20 \ln (.25) = -20 \ln (\frac 14) = 20 \ln (4) = 40 \ln 2$$
We have $$0.25=\frac 14=2^{-2}$$, so $$\log (0.25) = -2 \log (2)$$
• yes, of course, $ln(1/x) = -ln(x)$ and $ln(a^b) = b ln(a)$ thank you both !!! – velsd Sep 4 at 11:09