



 
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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