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| Math Central | Quandaries & Queries |
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Question from Katie, a student: Solve x=At^be^-ct for b in terms of c. |
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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