2
$\begingroup$

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?

$\endgroup$
  • 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
3
$\begingroup$

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 $f$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.