Minimum distance between $e^x$ and $\ln x$ 
How to find the minimum distance between the two curves $e^x$ and $\ln
 (x)$ ?

No idea how to find the common normal in this case.Help!
 A: Hint: 


*

*Each graph forms the boundary of a convex region.  The line segment of minimal distance between the two curves must therefore be unique.

*The picture is symmetric about the line $y = x$, so reflecting the line segment through this line must yield another line segment of minimal length.
We can deduce that the slope of the common normal must therefore be $-1$.
A: 
Minimum Distance between Two curve is Distance between two parallel tangents 
drawn at point $P$ and $Q$ on the curves.
and Here $f(x)=e^x$ and $f(x)=\ln(x)$ are Inverse of each other .
So it is Symmetrical about $y=x$ Line.
Let We take any point $P(x_{1},y_{1})$ on $f(x) = \ln(x)\;,$ Then Slope of tangent at $P(x_{1},y_{1})$ to the curve
$f(x)=\ln(x)$ is $\displaystyle f'(x_{1})=\frac{1}{x_{1}}$ and Slope of line $f(x)=x$ is $f'(x)=1$
Now If line $f(x)=x$ and Tangent at point $P(x_{1},y_{1})$ to the curve are parallel.
Then $\displaystyle \frac{1}{x_{1}} = 1\Rightarrow x_{1}=1$ Now We get $y_{1}=f_{x_{1}}=\ln(1) =0$
So we Get Coordinate of $P(1,0)$
Now Distance between Line $x-y=0$ from $P(1,0)$ is $\displaystyle = \frac{1}{\sqrt{2}}$
So $PQ = 2\times $ of Distance of line from $\displaystyle P(1,0) = \frac{2}{\sqrt{2}}=\sqrt{2}$ Unit.
