# Solve for $x$: $2^x = x^3$

What category of equation is this?

What methods are available to solve it?

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

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According to intermediate value theorem, one root of $f(x)=2^x-x^3$ is between $1$ and $3$. – Babak S. Jun 27 '12 at 11:48
Since it is not equivalent to a polynomial equation, it would be called a transcendental equation. BTW, that Wikipedia article needs some expert cleanup. – hardmath Jun 27 '12 at 13:08

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 $-\log 2/3$:

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

Hence:

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

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.

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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 $2.$

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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. – hardmath Jun 27 '12 at 13:03
+1 for nice answer. ;) – Babak S. Jun 27 '12 at 16:59
• 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
• ...
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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 $6.3$.

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Can you explian the third point please ? – Belgi Jun 27 '12 at 11:50
@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 $3$. – Marc van Leeuwen Jun 27 '12 at 11:54
3 Unique points. I love the way you attack the problem. – Babak S. Jun 27 '12 at 11:56