Show that there is a simultaneously tangent line to both the curves $y = e^x$ and $y = \ln x$

Question

I found the derivatives of both of the curves but I'm having trouble on how to move on from here:

$y = e^x\implies y' = e^x$

$y = \ln x\implies y' = \frac{1}{x}$

Answer 1

The tangent to $y=\ln x$ at $(1,0)$ does not intersect the curve $y=e^x$.

On the other hand, the tangent to $y = \ln x$ at $(10,\ln 10)$ is below $e^x$ at $(10, \ln 10)$ but above $e^x$ at $(0,\ln(10)-1)$, since $\ln(10)-1 \approx 1.303$. So that tangent does intersect $e^x$ at least once.

Thus, if you slide a tangent smoothly along $y=\ln x$ such that its point of tangency moves from $(1,0)$ to $(10,\ln(10))$, at some point the number of intersections between the tangent and $e^x$ will change from "none" to "some".

Hopefully you can argue that in that particular position, the tangent to $y=\ln x$ is necessarily also a tangent to $y=e^x$. (Precisely how to conduct that argument depends on the level of rigor expected of you in the context you're given the exercise; you might be able to get away with greater or smaller amounts of handwaving).

Note that the task you quote doesn't ask you to find the common tangent, just to prove that a common tangent exists somewhere.

Answer 2

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}

Answer 3

Let the tangent have the equation $y=ax+b$.

The condition of tangency to the exponential at $x_e$ is written

$$ax_e+b=e^{x_e},a=e^{x_e}$$

and that to the logarithm at $x_l$ is

$$ax_l+b=\log x_l,a=\frac1{x_l}.$$

We eliminate $x_e,x_l$,

$$a\log a+b=a$$

$$1+b=-\log a$$

then $b$, and we get

$$a\log a-a-\log a-1=0.$$

As the LHS is continuous, negative at $a=e$ and positive at $a=e^2$, there is certainly a solution, hence a common tangent.

---

As mentioned by @user3733558, there is another solution. By symmetry, it must correspond to the substitution $a\leftrightarrow\dfrac1a$. Indeed we verify

$$-\frac{\log a}a-\frac1a+\log a-1=\frac{a\log a-a-\log a-1}a=0.$$