I'm trying to solve a bigger problem however I am stuck at this step:

How can I solve:

$$ 2^x - x = 5 $$

any hints/tips/steps please?

  • 1
    $\begingroup$ Trial and error? 1 doesn't work, 2 doesn't work, ... (If you plot $y=2^x$ and $y=5+x$ in the same diagram, you'll see that there are two solutions, but I don't think the second one has a simple closed form.) $\endgroup$ – Hans Lundmark Oct 23 '11 at 13:42
  • $\begingroup$ Normally I might say something about Newton's method or about attractive fixed points. But in this case the answer is staring you in the face. $\endgroup$ – Michael Hardy Oct 23 '11 at 13:49
  • 1
    $\begingroup$ One solution is "obvious". The other real solution needs the services of Lambert. $\endgroup$ – J. M. is a poor mathematician Oct 23 '11 at 13:53
  • $\begingroup$ the second solution is somewhere between $-5$ and $-4$ $\endgroup$ – Peđa Terzić Oct 23 '11 at 14:22
  • $\begingroup$ $\approx -4.969$ $\endgroup$ – Peđa Terzić Oct 23 '11 at 14:57

As alluded to in the comments there is an integer solution. For the other solution is existence of a solution good enough? You can use the Intermediate Value Theorem on the function $f(x)=2^x-x-5$. It is negative at $x=0$ and positive at $x=-6$. So, somewhere in between the IVT says there must be a $0$. Or you can use Newton's method on $f$ to approximate the $0$ of $f$.


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