Value of $k$ for which $e^x = kx$ has $1$ solution I need to work out the value of $k$, where $k&gt;0$, for which $e^x=kx$ has $1$ solution. I've done it somewhat intuitively as follows: 


  $e^x=kx$
  
  By inspection we can see that when $x=1$, the exponent in the LHS and
  the multiplier(?) in the RHS are irrelevant, leaving us with
  
  $e=k$
  
  Therefore $e^x=kx$ has $1$ solution at $x=1$.


This is not a very rigorous solution at all though. What is a more proper way to solve this?
The idea is that the two functions $f(x)=e^x$ and $g(x) = kx$ touch only once; that is, $g$ is tangent to $f$ at some point.

Denote the tangent point $x_0$. Then $f(x_0)=e^{x_0}=kx_0=g(x_0)$. The slopes of the two functions have to be the same at this point as they are tangent:
$$f'(x_0) = g'(x_0)$$
$$e^{x_0} = k$$
and so
$$ kx_0 = k$$
The only solution is $x_0=1$ and the value of $k$ is then $k=e^{x_0} = e$.

Postscript: Negative $k$

I assumed above that $k$ is positive. For $k$ negative we can observe that
$$
  \lim_{x \rightarrow -\infty} (e^x - kx) = -\infty
$$
and
$$
  \lim_{x \rightarrow +\infty} (e^x - kx) = +\infty
$$
so that by the intermediate value theorem there exists a point at which $e^x-kx=0$. The function $h(x)=e^x-kx$ is monotonically increasing, as $h'(x)=e^x-k&gt;0$, so there is only one point at which $e^x=kx$.
It can be shown that for all $0&lt;k&lt;e$, $e^x=kx$ has no solution. (This is done using calculus, by showing that $e^x-kx$ has a minimum of $k-klog(k)$. Since $0&lt;k&lt;e$, we know that $log(k)&lt;1$, hence $k-klog(k)&gt;0$)

If $k&gt;e$, we know that $e^x-kx$ will be negative for some $x$ (At $x=log(k)$). Since, $e^x-kx$ grows without bound as $x$ approaches infinity or negative infinity. We can show that $e^x-kx$ crosses the x axis twice (a contradiction)