Changing an exponential function to logarithmic

Question

I have a question stating that $P=75e^{-0.005t}$ and they want to get t by itself.

I used the example $y=2^x = x=log_2(y)$

To find that $-0.005t = 75ln(P)$

So $t=\frac{75ln(P)}{-0.005}$

However apparently this isn't correct. Can someone please show me where I went wrong (not so much another way of doing it, but what was wrong in the working I used)

Thanks

Answer 1

$$\begin{align} P=75e^{-0.005t} & \iff \dfrac P{75} = e^{-0.005t} \\ \\ &\iff \ln\left(\frac P{75} \right) = \ln\left(e^{-0.005t}\right) \\ \\ & \iff \ln\left(\frac P{75} \right) = -0.005t \\ \\ & \iff t = -\left(\frac{\ln P - \ln (75)}{0.005}\right)\end{align}$$

Answer 2

P=75exp(-0.005t) ===> ln(P) = ln(75) + ln(exp(-0.005t)) ===> ln(P) = ln(75) - 0.005t ===> t= (ln(P) - ln(75))/-0.005