Find the minimum value of k $(k \in I)$ for which the equation $e^x =kx^2$ has exactly three real solution. Problem : Find the minimum value of k $(k \in I)$ for which the equation $e^x =kx^2$ has exactly three real solution. 
My approach : 
We apply log on both sides $x=2\ln(k x^2)$
$\Rightarrow \frac{x}{2} =\ln kx$ 
$\Rightarrow \frac{x}{2} =\ln k +\ln x$
I am not getting any clue further also whether this is right approach or not please suggest it will be of great help. Thanks.  
 A: For $k\le 0$, there will be no real solution at all for $kx^2\le 0<e^x$.
For $k>0$ there will always be exactly one solution with $x\le 0$.
For small positive $k$, there is no positive solution, for suitable $k$ the curves will touch and for larger $k$ we have two positive real solutions (so three real solutions in total).
We ned to find the boundary case that the curves touch, i.e., find $k>0$ and $x_0>0$ such that the functions and derivative agree at $x_0$:
$$ e^{x_0}=kx_0^2\qquad \text{and}\qquad e^{x_0}=2kx_0$$
We conclude $x_0=2$, hence $$\tag1k=\frac{e^2}4.$$
The value given by $(1)$ is the infimum of all $k$ for which there are exactly three distinct real solutions. If we count the tangential case as a solutoin of multiplicity two, then the value given by $(1)$ is also the minimum value of $k$ forwhich there are (with multiplicity) three real solutions.
A: Why not to look at function $$f(x)=\frac{e^x}{x^2}$$ Its derivative $$f'(x)=\frac{e^x (x-2)}{x^3}$$ cancels for $x=2$ and $f(2)=\frac{e^2}{4}$ which is a minimum by the second derivative test.
So, three real roots as soon as $k>\frac{e^2}{4}$ (two are positive and one is negative).
