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.
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.
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:
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: