How do you solve $x^2 = \left(\frac 12\right)^x$?

I'm having trouble finding the steps to solve for $x$. The solutions to this equation are $x=-4$, $x=-2$, and $x=0.76666$ when solved graphically and through the solve function of a TI-nspire cx CAS.

I tried to isolate $x$ by using various log and power rules, but the result was still something I did not know how to solve.

$$x^2 = \left(\frac 12\right)^x$$

$$\log x^2=\log\left(\frac12\right)^x$$

$$2\log x=x\log\left(\frac12\right)$$

$${\log x \over x}={\log \frac12 \over 2}$$

I also tried the following:

$$x^2 = \left(\frac 12\right)^x$$

$$x^2 = {1^x \over 2^x}$$

$1^n=1$ for all real $n$

$$x^2={1 \over 2^x}$$

$$x=\sqrt {1 \over 2^x} ={1 \over \sqrt {2^x}}={1 \over [2^x]^{1/2}}= {1 \over 2^{x/2}}$$

How do you do this?

• You'll need the en.wikipedia.org/wiki/Lambert_W_function. Apr 30, 2016 at 2:40
• It's a tough one to solve. You can possibly do it using the Lambert-W function (if you've heard of it). Apr 30, 2016 at 2:41
• I have never heard of it before. I am currently taking AP Calculus BC Apr 30, 2016 at 2:45
• What these comments are getting at is that there is no way to solve this equation in general with the usual pre-calculus methods. It's not that you haven't stumbled upon the correct way to rearrange the equation, it's that there is no such way. Apr 30, 2016 at 2:45
• With your result $x=2^{-x/2}$ you can find the solutions by recursion after a few iterations. For example, if you take your initial guess $x=1$ you will converge to the solution $x=0.7666$. Apr 30, 2016 at 2:49

Hint: $$x^2=\left(\frac{1}{2}\right)^x\implies x=\sqrt{\left(\frac{1}{2}\right)^x}\implies x=\sqrt{\left(\frac{1}{2}\right)^\sqrt{\left(\frac{1}{2}\right)^\sqrt{\left(\frac{1}{2}\right)^\sqrt{\left(\frac{1}{2}\right)^{........}}}}}$$

• Could you please explain how $x=\sqrt{\left(\frac{1}{2}\right)^x}\implies x=\sqrt{\left(\frac{1}{2}\right)^\sqrt{\left(\frac{1}{2}\right)^\sqrt{\left(\frac{1}{2} \right)^\sqrt{\left(\frac{1}{2}\right)^{........}}}}}$?
– Ovi
Aug 3, 2016 at 16:20
• Just put the value of $x$ in the r.h.s and put again.. Aug 30, 2017 at 3:36

You can use Lambert $W$, the inverse to $x\mapsto xe^x$: \begin{align} x^2 &=\frac{1}{2^x}\\ x^22^x &=1\\ x^2e^{x\ln(2)} &=1\\ xe^{x\ln(2)/2} &=\pm1\\ x\ln(2)/2 e^{x\ln(2)/2} &=\pm\ln(2)/2\\ W\left( x\ln(2)/2 e^{x\ln(2)/2}\right) &=W\left(\pm\ln(2)/2\right)\\ x\ln(2)/2 &=W\left(\pm\ln(2)/2\right)\\ x &=\frac{2W\left(\pm\ln(2)/2\right)}{\ln(2)}\\ x &\approx0.766664\ldots,-2,\text{ or }{-4} \end{align}

In the last line there are three results, firstly owing to the $\pm$ in the input to $W$, and secondly owing to the second branch of $W$ which admits small negative inputs.

You're forgetting that $x$ can also be less than $0$. Consider the function $$f(x)=\log(x^2)+x\log2=2\log|x|+x\log2$$ (natural logarithm), defined for $x\ne0$.

We can easily compute the relevant limits: $$\lim_{x\to-\infty}f(x)=-\infty, \quad \lim_{x\to0^-}f(x)=-\infty, \quad \lim_{x\to0^+}f(x)=-\infty, \quad \lim_{x\to\infty}f(x)=\infty$$ We also have $$f'(x)=\frac{2}{x}+\log2$$ which is positive for $x>0$. The function has a local maximum at $-2/\log2$ and $$f(-2/\log2)=2\log\frac{2}{\log2}-\frac{1}{\log2}>0$$ Thus the equation $f(x)=0$ has three solutions, two in the interval $(-\infty,0)$ (that are obviously $-4$ and $-2$) and one in the interval $(0,\infty)$.

Since $f(1)=\log2>0$ and $f(1/2)=-\frac{3}{2}\log2<0$, we know that this solution is in $(1/2,1)$.

You can try an iteration, setting $x_0=1$ and $x_{n+1}=(1/2)^{x_n/2}$ \begin{align} x_0&=1 \\ x_1&=0.70710678118654752441 \\ x_2&=0.78265402735568027271 \\ x_3&=0.76242798854892438203 \\ x_4&=0.76779124029202027521 \\ x_5&=0.76636542509760295727 \end{align} and so on; this seems to converge quite fast.

There is no way to express the solution other than using the Lambert function trick, or other similar ones.

HINT.-A clear solution is $x=-2$ and there is another in the open interval $]0,1[$ which you can try several ways approximate (no elementary methods to get directly the solution because it is what some people called a trascendental equation).

The single positive solution by be found to any level of precision by recursion on the relation $x=2^{-\frac{x}{2}}$ letting $x_1=1$. Then $$x_2=2^{-\frac{1}{2}}\approx 0.707106781$$ $$x_3=2^{-\frac{x_2}{2}}\approx 0.782654027$$ until one arrives at $$x_{19}=2^{-\frac{x_{18}}{2}}\approx 0.766664696$$ For $N\ge19$ to 9 decimal precision, $x_N\approx 0.766664696$.