<s>[INST] Show that there is a simultaneously tangent line to both the curves $y = e^x$ and $y = \ln x$ I found the derivatives of both of the curves but I'm having trouble on how to move on from here:
$y = e^x\implies y' = e^x$
$y = \ln x\implies y' = \frac{1}{x}$
 [/INST] Given  \begin{align} y = e^x\implies y' = e^x$, $y = \ln x\implies y' = \frac{1}{x}\end{align}
Tangent line to $y = e^x$ at x=a
\begin{align} y=e^ax+e^a(1-a)\end{align}
Tangent line to $y=lnx$ at x=b\begin{align} y=\frac{1}{b}x+\ln b-1\end{align}
When the two lines become one line:
\begin{align} e^a=\frac{1}{b}, e^a(1-a)=\ln b-1\end{align}
Solve the above equations about a and b, \begin{align}b \approx 0.124, a\approx ln(\frac{1}{0.214}) \\b\approx 4.681, a\approx ln( \frac{1}{4.681}) \end{align}
The common tangent lines  \begin{align}  y= \frac{1}{0.214}x+\ln(0.214)-1 \\y= \frac{1}{4.681}x+\ln(4.681)-1  \end{align}
</s>