Question

I have to put f(x) = 37.29ℯ^(-0.69 x) into a logarithmic function? I know that y = b^x is the same as x = logby, but I don't understand. Will someone walk me through this?

Answer 1

Have you considered simply introducing the logarithm?

\(y = 37.29e^{-0.69x} \longrightarrow \log(y) = \log(37.29) - 0.69x\) You can solve for x if you like.

Answer 2

could you explain how you did that? i am so, so confused. Where did the e go?

Answer 3

I just introduced the logarithm. i didn't do anything else.

Maybe this part?
\(\log\left(e^{-0.69x}\right) = -0.69x\log(e) = -0.69x\)