Math CentralQuandaries & Queries


Question from Katie, a student:

Solve x=At^be^-ct for b in terms of c.
The e is Euler's number, not a letter or random placeholder.
I've tried 2 methods on my calculator, and I am unsure as to why they give me different answers, and which one is correct.
For some reason, when placing a multiplication sign in between the A and the first t, I get a completely different answer. Why is this?
Also, which is the right answer for the question I have?
Thank you!
Katie :)

Hi Katie,

I can't help you with your calculator difficulties but I can show you how I would approach this problem. I would take the natural logarithm (log) of both sides using the facts that

\[\log(P \times Q) = \log(P) + \log(Q)\]


\[\log\left(P^Q\right) = Q \times \log(P).\]

Thus, taking the log of both sides of

\[x=A t^b e^{-ct}\]

I get

\[\log(x) = \log\left(A t^b e^{-ct}\right) = \log(A) + \log\left(t^b\right) + \log\left(e^{-ct}\right) = \log(A) + b \log(t) + (-ct) \log(e) = \log(A) + b \log(t) -ct.\]

Now solve for $b$ in terms of $c.$


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