Solving simultaneous equation involving quadratic and exponential

Question

I have two equations $y=x^2$ and $y=e^{x+1}$. I want to solve them simultaneously and find $x.$ But I am having trouble solving them. I have tried the following: \begin{aligned}x^2&=e^{x+1}\\ \ln(x^2)&=x+1\\ 2\ln(x)&=x+1\\ \ln(x)&=\frac{x+1}{2} \end{aligned} and then I have no more ideas.

Answer 1

For $x\le0$, the function $x^2$ is strictly decreasing, and the function $e^{x+1}$ is strictly increasing, so there is at most one solution. By inspection $x=-1$ is a solution, so there are no others for $x<0$.

For $x>0$ we want to show that $$e^{x+1}>x^2\quad(*)$$ so that there are no solutions for $x>0$. That would be easy with calculus, but the OP has classified it as precalculus, so we have to work a little harder.

Note first that $(x-1)^2>2$ for $x\ge3$, so (rearranging) $2x^2>(x+1)^2$. We claim that for any positive integer $n>3$ we have $2^n\ge n^2$. It is true for $n=4$. Assume it is true for $n$, then $2^{n+1}=2\cdot2^n\ge2n^2>(n+1)^2$, so it is true for $n+1$ and hence for all $n$.

Now consider $4\le n<x<n+1$. We have $e^{x+1}>2^{x+1}\ge2^{n+1}\ge (n+1)^2\ge x^2$. So we have established (*) for $x\ge4$.

$(*)$ is obviously true for $0\le x\le 1$. It is true for $1\le x\le2$ because $e^{x+1}>2^{1+1}=4\ge x^2$. It is true for $2\le x\le 4$ because $e^{x+1}>2.7^3>16\ge x^2$.