How would one solve the equation $2^x=x^3-1$? 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.
 A: 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.
A: You can use pencil and paper to give a quick statement of:


*

*Whether there is a solution, and

*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:


*

*That there is a solution, and

*The consecutive positive integers between which the solution lies.

