# Solving $x^2 - 1 = e^x$

Can someone help me solve the equation $x^2 - 1 = e^x$ ?

I tried taking the natural logarithm of both sides but I don't know where to go from there..

I got:

$\ln(x^2 -1) = x$ But I don't know how to solve it from here. Any help please?

• It appears that $x\lt -1$... have you tried graphing the two sides as functions and looking for an intersection? – abiessu Nov 30 '13 at 17:53
• This is a transcendental equation. You can only solve it using a numerical method. – abnry Nov 30 '13 at 17:53
• Hint: you can draw $f=x^2-1$ and $g=e^x$. – Babak Miraftab Nov 30 '13 at 17:54
• You can also look at $y=-x-1\rightarrow y^2+2y=e^{-y-1}\rightarrow \ln(y+2)+\ln y=-y-1$... – abiessu Nov 30 '13 at 17:57

You first have to see where the solutions are. You can consider the function $$f(x)=e^x-x^2+1$$ and compute $$\lim_{x\to-\infty}f(x)=-\infty,\qquad \lim_{x\to\infty}f(x)=\infty.$$ Now we want to see whether the function has stationary points. The derivative is $$f'(x)=e^x-2x$$ and we want to see where it's positive and negative. From $$\lim_{x\to-\infty}f'(x)=\infty,\qquad \lim_{x\to\infty}f'(x)=\infty$$ we know that the derivative has a minimum, which will be attained where $f''(x)=0$. Since $$f''(x)=e^x-2,$$ we know it's zero at $\log 2$. Now $$f'(\log 2)=2-2\log2=2(1-\log 2)>0$$ because $2<e$ so that $\log 2<1$.
Thus we know that $f'$ has a positive minimum, so we can argue that $f'(x)>0$ for all $x$. As a consequence, $f$ is increasing, so it assumes the value $0$ only once.
Since $f(-1)=e^{-1}<0$ and $f(-2)=e^{-2}-3<0$ (because certainly $e^{-2}<1$), we know that the unique solution $a$ of your equation satisfies $-2<a<-1$.
With some numeric method you can get better approximations of $a$.