Why cant I solve for t here?

Question

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 $t$ ? $$20=20 + 80e^{-kt}$$

Answer 1

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 $k=\frac{1}{20}\ln \frac{8}{7}>0$.

Answer 2

$$20=20+80e^{-\text{k}t}\Longleftrightarrow\ln\left(\frac{20-20}{80}\right)=-\text{k}t$$

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 $t\to\infty$

Answer 3

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.

Answer 4

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.