# 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

• Using desmos.com I can easily check the point of intersection but how to verify it without using desmos.com – Samar Imam Zaidi Jun 10 at 15:15
• For different $a$ the curve has different points of intersection. – Vedant Chourey Jun 10 at 15:24
• But is there any way to find it – Samar Imam Zaidi Jun 10 at 15:28
• You can fix one value of $a$ , after some hit and trial you probably get solutions – Vedant Chourey Jun 10 at 15:33
• Looking at the curves only the domain $[0, \infty)$, you can realize that there will always be an intersection for any given value of $a$. So, there are infinite values of $a$ that will produce an intersection with $x^2$ in $[0, \infty)$. In the negative part, you may not always observe an intersection, as it will depend on the value of $a$ and how fast it grows. – Dunkel Jun 10 at 15:44

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 $$b < b_0$$.

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

So we have at least 2 solutions when $$1.

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 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 :

$$u^2/2+(1-2/\ln{a})u+1=0$$

$$u={(2/\ln{a} -1)+\_\sqrt{[2/\ln{a}-1]^2-2}}$$