How do we solve the equation $2^{x^2-3}=x^{-1/3}$ algebraically?

Question

This question was from Khan Academy and, even though Sal solved it through graphing, I want to know how it can be solved algebraically.

Here are the steps that I have tried:

$2^{x^2-3}=x^{-1/3}$

$2^{{(x^2-3)}^{-3}}=x^{{(-1/3)}^{-3}}$

$2^{-3x^2+9}=x$

$\log_2(x)=-3x^2+9$

After this step I do not know what to do.

Answer 1

It's not always possible to solve equations algebraically. This one defies elementary methods.

In the "real world" we'd use a numerical approximation method to solve the equation.

Answer 2

The solution is $$ x = \sqrt{\frac{W(1572864 \ln(2))}{6 \ln(2)}} $$ where $W$ is the Lambert W function.

EDIT: To see this, let $y = 6 x^2 \ln(2)$, and rewrite the equation as $$ \frac{e^{y/6}}{8} = \frac{2^{y/(6 \ln 2)}}{8} = y^{-1/6} (6 \ln(2))^{1/6}$$ Taking the $6$'th power of each side and multiplying by $8^6 y$, it becomes $$y e^y = 1572864 \ln(2)$$ so $y = W(1572864 \ln(2))$.

Answer 3

There is no explicit solution if you cannot use Lambert function and some numerical method will be required.

Solving $$2^{x^2-3}=x^{-1/3}$$ is just the same as finding the zero of function $$f(x)=\log \left(2^{x^2-3} \sqrt[3]{x}\right)$$ If you plot it, you will see that the root is close to $1.65$ which is close to $\sqrt 3$.

Developing $f(x)$ as a Taylor series built around $x=\sqrt 3$ would give $$f(x)=\frac{\log (3)}{6}+\frac{ (1+18 \log (2))}{3 \sqrt{3}}\left(x-\sqrt{3}\right)+\left(\log (2)-\frac{1}{18}\right)\left(x-\sqrt{3}\right)^2 +O\left(\left(x-\sqrt{3}\right)^3\right)$$ which is a quadratic equation in $\left(x-\sqrt{3}\right)$. Just solve it and pick the closest root.

Converted to decimal, this would give $x\sim 1.660183$ while the exact solution is $x=1.660186\phantom{for some reason edits must be 6 characters.}$.