Value of $k$ for which $e^x = kx$ has $1$ solution I need to work out the value of $k$, where $k>0$, 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?
 A: 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 $h'(x)=e^x-k>0$, so there is only one point at which $e^x=kx$.
A: Since $e^x>0$ for all real $x$, then $y=e^x$ does not intersect $y=kx$ at all if $k=0$.
If $k<0$, then observe that $e^x-kx$ is strictly increasing, with $$\lim_{x\to\infty}(e^x-kx)=\infty$$ and $$\lim_{x\to-\infty}(e^x-kx)=-\infty.$$ From the strict monotonicity, there is at most one solution to $e^x=kx$, and from an application of the Intermediate Value Theorem, there is at least one solution.
Suppose that $k$ is positive and that $y=kx$ intersects $y=e^x$ at exactly one point--equivalently, that $f(x)=e^x-kx$ has exactly one zero. Now, $f'(x)=e^x-k$, and by observing the sign of $f'(x)$, we conclude that $f$ is decreasing on $(-\infty,\ln k)$ and increasing on $(\ln k,\infty)$, achieving a global minimum when $x=\ln k$. Noting that $f(x)\to\infty$ as $x\to\pm\infty$, it follows that the minimum value of $f(x)$ cannot be negative, for otherwise, $f(x)$ would have two zeroes--one in $(-\infty,\ln k)$ and one in $(\ln k,\infty)$--but on the other hand, the minimum value of $f(x)$ cannot be positive, either, for otherwise $f(x)$ would have no zeroes. Thus, the minimum value of $f(x)$ (which, recall, is achieved at $x=\ln k$) is $0$, meaning  $$0=f(\ln k)=e^{\ln k}-k\ln k=k-k\ln k=k(1-\ln k).$$ Since $k>0$, this means that $1-\ln k=0$, so $\ln k=1$, and so $k=e$.
A: You want the two curves to be tangent so the slope at the common point is the same.  This requires a common solution to $e^x=kx$ and $e^x=k$.  Equating the right sides, we get $kx=k$, so $x=1$ or $k=0$, but $k=0$ is not allowed, so $x=1, k=e$
A: An alternative to the methods presented so far is to explicitly solve the equation for $x$ in terms of the Lambert W function, $\text{W} (x)$.
Rewriting the equation as 
$$-x e^{-x} = -\frac{1}{k},$$
on solving for $x$ we have
$$x=−\text{W}_\nu \left (−\frac{1}{k} \right ).$$
Here $\nu$ denotes the branch of the Lambert W function.
As $k > 0$, two distinct real roots occur when $0 < k < e$. These roots are given by $−\text{W}_0(−1/k)$ and $−\text{W}_{−1}(−1/k)$. Here $\text{W}_0(x)$ denotes the principal branch of the Lambert W function while $\text{W}_{−1}(x)$ denotes its secondary real branch. 
For one real root only, this occurs at the branch point between the two real branches $\text{W}_0 (x)$ and $\text{W}_{-1} (x)$, it occurring when the argument for the Lambert W function is equal to $-1/e$. Thus $k = e$ and as
$$\text{W}_0 \left (-\frac{1}{e} \right ) = \text{W}_{-1} \left (-\frac{1}{e} \right ) = -1,$$
we find it occurs at the point $x = 1$.   
A: It can be shown that for all $0<k<e$, $e^x=kx$ has no solution. (This is done using calculus, by showing that $e^x-kx$ has a minimum of $k-klog(k)$. Since $0<k<e$, we know that $log(k)<1$, hence $k-klog(k)>0$)
If $k>e$, 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)
