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I am stuck on this problem, and the book says that $kt=\ln(2)$ is the correct answer.

I am working with Newton's law of cooling, and I am supposed to solve for $kt$, supposing a $t$ that makes the temperature difference $(u_0-T)$ half of its original value. I have gotten this far: $$\frac{1}{2}(u_0-T)=(u_0-T)e^{-kt}+T\\\frac{1}{2}u_0-\frac{1}{2}T+T=(u_0-T)e^{-kt}\\ \frac{(u_0+T)}{2(u_0-T)}=e^{-kt}$$ And this is where I get stuck. I have tried doing this $$\ln\left(\frac{(u_0+T)}{2(u_0-T)}\right)=-kt\\ \ln(u_0+T)-\ln(2)-\ln(u_0-T)=-kt\\ \ln\left(\frac{(u_0+T)}{(u_0-T)}\right) - \ln(2) = -kt$$ But unless that first term becomes zero somehow, I don't know how to proceed. I would be very grateful if anyone could point me in the right direction!

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If the temperature difference is half of the original difference, then the temperature will be $T + {1 \over 2} (u_0 -T)$.

This gives $T+ e^{-kt} (u_0-T) = T + {1 \over 2} (u_0 -T)$ from which we get $e^{-kt} = {1 \over 2}$. The answer follows.

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  • $\begingroup$ Oh my gosh, such a stupid mistake on my part. Thank you! $\endgroup$ – riley lyman Jan 26 '17 at 2:27
  • $\begingroup$ @rileylyman: No problem, everyone makes such mistakefs. $\endgroup$ – copper.hat Jan 26 '17 at 2:27
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Where you have $\dfrac 1 2 (u_0-T)=(u_0-T)e^{-kt}+T,$ you need $\dfrac 1 2 (u_0-T) = (u_0-T)e^{-kt}.$

From that you get $\dfrac 1 2 = e^{-kt}.$

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