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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

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  • $\begingroup$ Using desmos.com I can easily check the point of intersection but how to verify it without using desmos.com $\endgroup$ – Samar Imam Zaidi Jun 10 at 15:15
  • $\begingroup$ For different $a$ the curve has different points of intersection. $\endgroup$ – Vedant Chourey Jun 10 at 15:24
  • $\begingroup$ But is there any way to find it $\endgroup$ – Samar Imam Zaidi Jun 10 at 15:28
  • $\begingroup$ You can fix one value of $a$ , after some hit and trial you probably get solutions $\endgroup$ – Vedant Chourey Jun 10 at 15:33
  • $\begingroup$ 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. $\endgroup$ – Dunkel Jun 10 at 15:44
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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 $b < b_0$.

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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.

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