Why cant I solve for t here? This is one of those cup of coffee questions. The room temperature is $20$ degrees.
$T = $ Temperature of Coffee, $t=$ time
How long until the coffee cools to room temperature if
$$T=20+80e^{-kt}$$
I have already found that $k=\frac{1}{2}\ln{\frac{8}{7}}$.
How to isolate $t$ ?
$$20=20 + 80e^{-kt}$$
 A: Logarithms:
$$T=20+80e^{-kt}\implies e^{kt}=\frac{80}{T-20}\implies kt=\log\frac{80}{T-20}\implies t=\frac1k\cdot\log\frac{80}{T-20}$$
Of course, the above is defined only for $\;T>20\;$ , so in your case there wouldn't be any solutions.
A: $$20=20+80e^{-\text{k}t}\Longleftrightarrow\ln\left(\frac{20-20}{80}\right)=-\text{k}t$$
But know we notice that $\frac{20-20}{80}=0$ and:
$$\lim_{x\to0}\ln(x)\to-\infty$$
So, there are no solutions (as a finite number) but when $\text{k}>0$, then $t\to\infty$
A: As you can see from below:$$\begin{align} & 20=20 + 80e^{-kt} \\ \implies & 80e^{-kt}=0  \\ \implies & e^{-kt}=0 \\ \implies & e^{kt} \to \infty \\ \implies & t \to \infty\end{align}$$
There is no unique solution to this equation. What you can conclude at most is that $t \to \infty$ since $k=\frac{1}{20}\ln \frac{8}{7}>0$.
A: With your model, the coffee actually never cools down to room temperature, because the equation
$$
20=20+80e^{-kt}
$$
is equivalent to the equation
$$
0=e^{-kt}.
$$
But because $e^x>0$ for all real numbers $x$, there is no $t$ that satisfies the equation above. 
