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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.

user376343
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Ashutosh Mishra
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    Not all the equations can be solved algebraically. Like $\tan x$ and $1/x$. Sometimes we just want the numbers of solutions of the equation which can be solved by graphing. – SarGe Jul 30 '20 at 16:19
  • Unlikely. Equations with $x$ both in an exponent and outside of it are usually not solvable for $x$ with elementary methods. Sometimes it's possible to use special functions like "Lambert-W", but here I doubt even that's possible. – Jair Taylor Jul 30 '20 at 16:20
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    Your best bet would probably be Newton's Method if you're familiar with basic calculus. – N. Bar Jul 30 '20 at 16:21
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    @AshutoshMishra please check my edit – user376343 Jul 31 '20 at 08:51

3 Answers3

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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 $y = W(1572864 \ln(2))$.

Robert Israel
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    I do not understand Lambert's W function since I am an Algebra student is there any way to solve this with Algebra. – Ashutosh Mishra Jul 30 '20 at 17:15
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    @AshutoshMishra No, $x$ is not an algebraic number, and $W$ is not an elementary function. – Robert Israel Jul 30 '20 at 19:34
  • @RobertIsrael Did You recognize you should use $y=6x^2\ln(2)$ at first or you just wanted to check the answer? I simplified $2^{x^2-3}=x^{-1/3}$ and obtained $\ln(x^2)+6x^2\ln(2)=18\ln(2)$ then I used the subsituation and get the same answer as yours. Did you do the same? – Aligator Jul 31 '20 at 15:11
  • I used Maple to get the answer first. – Robert Israel Jul 31 '20 at 15:34
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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.

Jolly Llama
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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 $x=1.660186\phantom{for some reason edits must be 6 characters.}$.

Jolly Llama
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Claude Leibovici
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