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?
 A: Hint:
$$x^2=\left(\frac{1}{2}\right)^x\implies x=\sqrt{\left(\frac{1}{2}\right)^x}\implies x=\sqrt{\left(\frac{1}{2}\right)^\sqrt{\left(\frac{1}{2}\right)^\sqrt{\left(\frac{1}{2}\right)^\sqrt{\left(\frac{1}{2}\right)^{........}}}}}$$
A: 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.
A: 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.
A: 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).
A: 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
\begin{equation}
x_2=2^{-\frac{1}{2}}\approx 0.707106781
\end{equation}
\begin{equation}
x_3=2^{-\frac{x_2}{2}}\approx 0.782654027
\end{equation}
until one arrives at
\begin{equation}
x_{19}=2^{-\frac{x_{18}}{2}}\approx 0.766664696
\end{equation}
For $N\ge19$ to 9 decimal precision, $x_N\approx 0.766664696$.
