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?

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

There is no analytical expression for solving this transcendental equation, not even in terms of the Lambert W function. The only solution would be using numerical algorithms, such as Newton's method. (The roots of its derivative can be expressed in terms of the Lambert W function, though).


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 $a$.


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