What category of equation is this?

What methods are available to solve it?

$2^x -x^3 = 0$ where $x\in\Bbb R$

  • $\begingroup$ According to intermediate value theorem, one root of $f(x)=2^x-x^3$ is between $1$ and $3$. $\endgroup$
    – Mikasa
    Jun 27, 2012 at 11:48
  • $\begingroup$ Since it is not equivalent to a polynomial equation, it would be called a transcendental equation. BTW, that Wikipedia article needs some expert cleanup. $\endgroup$
    – hardmath
    Jun 27, 2012 at 13:08

4 Answers 4


You can find a solution in terms of the Lambert W Function. Rewrite as:

$$ 1 = \frac{x^3}{2^x} = x^3 \exp(-x\log 2) $$

and take the real cube root:

$$ 1 = x \exp \left(-\frac{x\log 2}{3}\right) $$

Now multiply by $-\log 2/3$:

$$ -\frac{\log 2}{3} = -\frac{x\log 2}{3} \exp\left(-\frac{x\log 2}{3}\right) $$


$$ x = -\frac{3W_0\left( -\frac{\log 2}{3}\right)}{\log 2} $$

where $W_0$ is the principal branch of Lambert's W. The value is about 1.37.

There is another real root between 9 and 10, which is found on the second branch of the Lambert W function:

$$x = -\frac{3W_{-1}\left(-\frac{\log 2}{3}\right)}{\log 2}$$

and whose value is about 9.94.


Consider $f(n)=2^n-n^3$. $f(9)=-217$ and $f(10)=24$,therefore,there exists a root between $9$ and $10.$ You can solve $f(n)=0$ numerically using Newton-Raphson method taking $x_0=9$.

Also $f(1)=1$ and $f(2)=-6$, therefore there will be a root between 1 and 2.

For $n<1$, $f(n)$ is always positive and for $n>10$, $f(n)$ is always positive,so there are no more roots other than between $9$ and $10, 1$ and $2.$

  • $\begingroup$ For the root between 1 and 2, the fixed point iteration $x := 2^{x/3}$ seems to work well (though not as fast as Newton!), while between 9 and 10 the inverse iteration $x := 3 \log_2 x$ seems to work. $\endgroup$
    – hardmath
    Jun 27, 2012 at 13:03
  • it is an exponential equation or an exponential diophantine equation if only integers (or fractions) are allowed
  • you may find a real or imaginary solution :
    • using iterations starting with $x_0=2$ ($x_{n+1}=2^{x_n/3}$, the converse $x_{n+1}=\frac {3\ln(x_n)}{\ln 2}$ or faster Newton-Raphson iterations)
    • with the LambertW function (since $n e^{-\frac {n\ln 2}3}=1$ is wanted)
    • graphically
    • ...

This is an equation without solution, if $n$ is supposed to be integer. In order for $2^n=n^3$ to hold, $n$ should be

  • positive, since $2^n$ is,
  • a power of $2$, since $n^3$ would otherwise have an odd prime factor,
  • divisible by $3$ in order for $2^n$ to be a cube.

Obviously the last two conditions are contradictory.

Since $n$ has now been renamed $x$ and has become real, there are more solutions. Probably just the two indicated by @avatar, but some effort is required to show that this is all, since $f: x\to 2^x-x^3$ is not a convex function: it has two inflexion points, one near $0.08$ and another one near $6.3$.

  • $\begingroup$ Can you explian the third point please ? $\endgroup$
    – Belgi
    Jun 27, 2012 at 11:50
  • 1
    $\begingroup$ @Belgi: $2^n$ has an obvious prime factorization (or should I say, it is a prime factorization), and prefect cubes are precisely those integers in whose prime factorization every prime factor occurs with multiplicity divisible by $3$. $\endgroup$ Jun 27, 2012 at 11:54
  • 1
    $\begingroup$ 3 Unique points. I love the way you attack the problem. $\endgroup$
    – Mikasa
    Jun 27, 2012 at 11:56

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