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.

  • 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

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.


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.

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