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 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

$\log(P \times Q) = \log(P) + \log(Q)$

and

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

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 $c.$

Penny

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