Differential Equations Simple Question

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!

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.
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 $\dfrac 1 2 = e^{-kt}.$