3
$\begingroup$

How would one solve the equation $2^x=x^3-1$?

I can't figure it out.

I managed to solve the easier $x^3 = 2^x$ using super-roots and the Lambert W function, but I can't seem to figure out how to solve it.

$\endgroup$
  • 4
    $\begingroup$ I'm afraid you need some numerical method. $\endgroup$ – Arnaldo Jan 14 '17 at 2:28
  • $\begingroup$ Do you have reason to believe there is a closed form? $\endgroup$ – Thomas Andrews Jan 14 '17 at 3:48
  • $\begingroup$ @ThomasAndrews Not at all; I should probably mention that I mostly just mean getting something like $x = \textrm{something containing no other } x\textrm{'s}$, no matter how ugly it looks :P $\endgroup$ – goose121 Jan 14 '17 at 4:43
  • $\begingroup$ The equation $2^x=ax^3-1$ is soluble for $a\approx .259$. I can tell you the solution, but I don't know how to get a closed value for $a$. The solution is $x = \frac{W(3e^{-3}) + 3}{\ln{2}}$. If this is useful I can write up an answer. $\endgroup$ – Χpẘ May 4 '17 at 19:40
1
$\begingroup$

First, I am assuming that you only want real roots of $2^x=x^3-1 $.

If $x < 1$, then $x^3-1 < 0$, and $2^x > 0$, so no negative roots.

Let $f(x) = 2^x-x^3+1 $.

According to Wolfy, $f(x) > 0$ for $0 < x < 1.58833$, $f(x) < 0$ for $1.58833 < x < 9.93693$, and $f(x) > 0$ for $9.93693 < x$.

$f'(x) =\ln 2\ 2^x - 3x^2 $.

According to Wolfy, which uses the Lambert W function, the positive roots of this are $x_1 = -\dfrac{2 W\left(-\frac{\log^{3/2}(2)}{2 \sqrt{3}}\right)}{\log(2)} \approx 0.589665 $ and $x_2 = -\dfrac{2 W_{-1}\left(-\frac{\log^{3/2}(2)}{2 \sqrt{3}}\right)}{\log(2)} \approx 8.1768 $.

If $x > x_2$, then $f'(x) > 0$. Therefore, if $x > 9.93693$, $f(x) > 0$.

You can argue using the higher derivatives of $f(x)$ to get more elementary methods of showing when $f(x) = 0$, but I will leave it at this.

$\endgroup$
0
$\begingroup$

You can use pencil and paper to give a quick statement of:

  1. Whether there is a solution, and
  2. The range the solution(s) must fall in if it/they exist.

First note that $2^x$ is positive for all $x$, and $x^3-1$ is negative for negative $x$. So if there is a solution, it must be non-negative.

Next note that both the RHS and LHS of the equation are strictly increasing for positive $x$. This simplifies things drastically.

Now make a table for the LHS and RHS of the equation for small values of $x$. More precisely, let's define two functions: $$f(x)=2^x$$ $$g(x)=x^3-1$$ and make tables for each function, for small non-negative inputs. (Actually do this; it's very simple.)

Observe which is greater, $f(x)$ or $g(x)$, for each value of $x$ that you check. Observe where the comparison changes (and observe that it changes).

This will tell you:

  1. That there is a solution, and
  2. The consecutive positive integers between which the solution lies.
$\endgroup$

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.