Curves $y=a^{x}$ and $y=x^2$. Let us assume two curves $y=a^{x}$ and $y=x^2$. Let  us assume that a>0. Values range of a for which curves have one solution, two solution and three solution.
Is there a way to check the values. I can draw both the curves but how to check the number of points of intersection
 A: Fix $a > 1$ (for $0 < a < 1$, we can set $x = -x$). For $x < 0$, one function is increasing and another is decreasing, thus there is exactly one root. For $x > 0$,
$$a^x = x^2 \Leftrightarrow
x \ln a = 2 \ln x \Leftrightarrow
\frac {\ln x} x = \frac {\ln a} 2.$$
$\ln(x)/x = b$ with $b > 0$ has no roots when $b > b_0 = 1/e$, one root when $b = b_0$ and two roots when $b < b_0$.
A: I think I have something. By the given hypothesis, $a>0$. 
METHOD 1
$$x^2-a^x =(x-a^{x/2})(x+a^{x/2})=0$$
So if $x-a^{x/2}$ has a positive maximum and goes to negative infinity for large absolute values of $x$, then we have 2 solutions. We have a single solution when that peak is zero. 
What is the maximum of $y=x-e^{x\ln{a}/2}$
The $x$ satisfying $0=1-\frac{\ln{a}}{2}e^{x \ln{a} /2}$
So: $x_0=\frac{2}{\ln{a}}\ln{\frac{2}{\ln{a}}}$
$$y(x_0)=\frac{2}{\ln{a}}\ln{\frac{2}{\ln{a}}}-\frac{2}{\ln{a}}$$
Note that $y(x_0)$ is zero when $a=e^{2/e}$
Now we want to find $y(x+c)=0$, sl
$$0=\frac{2}{\ln{a}}\ln{\frac{2}{\ln{a}}}+c-e^{\frac{\ln{a}}{2}(\frac{2}{\ln{a}}\ln{\frac{2}{\ln{a}}}+c)}$$
or:
$$0=\frac{2}{\ln{a}}\ln{\frac{2}{\ln{a}}}+c-\frac{2}{\ln{a}}e^{\frac{c\ln{a}}{2}}$$
Using Taylor Series to second order:
$$0=\frac{2}{\ln{a}}\ln{\frac{2}{\ln{a}}}+c-\frac{2}{\ln{a}}(1+\frac{c \ln{a}}{2}+\frac{c^2  \ln{a}^2}{8})$$
Cleaning up and rearranging:
$$c^2=\frac{8\ln{\frac{2}{e\ln{a}}}}{(\ln{a})^2}$$
METHOD 2
$$a^x=x^2$$
when
$$x\ln{a}=2\ln{x}$$
which implies 
$$x/\ln(x)=2/\ln{a}$$
$x/\ln{x}$ takes on all negative real values, so we have at least one solution for all $a$ between $0$ and $1$. But if $a>1$, by symmetry, this implies a solution in the negative reals , i.e. $a^{-x}=x^2$ satisfies our equation for some real x less than zero. 
$x/\ln{x}$ can only take on positive values greater than $e$ since that is the global minimum in the positive reals. 
When $a>1$, $\ln{a}>0$. We need to satisfy :
$$2/\ln{a}>e$$
So $\ln{a}<2/e$ or, equivalently,  $a<e^{2/e}$
So we have at least 2 solutions when $1<a<e^{2/e}$. 
We transition from having a single solution to having two only at a point of tangency. 
$a^x=x^2$ and
$x \ln{a}=2x$
This is only satisfied when $x=2/\ln{a}$ for some $a$. 
To be at an intersection of $x^2$ and $a^x$, we need $x/\ln{x}=2/\ln{a}$ but at the point of single intersection, the point of tangency, we also need $x=2/\ln{a}$.
But $x/\ln{x}=x$ only when $x=e$. 
So $e=2/\ln{a}$. This implies $\ln{a}=2/e$, then $a=e^{2/e}$. 
So $(e,e^2)$ with $a=2^{e/2}$ is our only value of $a$ for which there are only two solutions. Otherwise we have 1 or 3. 
We have 3 solutions for $1<a<e^{e/2}$ with positive real solutions coming in pairs except at the one condition just specified. 
$x/\ln{x}=2/\ln{a}$ has a solution when $e^u/u=2/\ln{a}$ (letting $x=e^u$) has a solution for u. 
We can approximate this using Taylor series and get a quadratic equation:
$$1+u+u^2/(2!)+u^3/(3!)+... = (2/\ln{a})u$$
or :
$$u^2/2+(1-2/\ln{a})u+1=0$$
Using the quadratic formula:
$$u={(2/\ln{a} -1)+\_\sqrt{[2/\ln{a}-1]^2-2}}$$
Should be a decent approximation. 
