How can I solve $x^{x^4}=64$?

Question

Once I google it, I got the result from here, where the real solutions are $x={\pm}2^{3/4}$. But it doesn't show how to get the result.

I also don't know what the function W mean in other solutions like $x=e^{1/4W(8i{\pi}n+24\log(2))}$.

Answer 1

Taking logarithm base 2 yields:

$$x^4\log_2x=6$$ Now, let $t=\log_2x$ then $x=2^t$ which implies $t16^t=6$ ie: $16^t=\frac{6}{t}$

Note that : $f(t)=16^t$ is always positive (implies $t>0$) and $f(t)$ is strictly increasing function. So, when $t>0$ function $g(t)=\frac{6}{t}$ is strictly decreasing function. So, if $f(t)=g(t)$ has root, then it is unique. By try and error we can find $t=\frac{3}{4}$ then $\log_2x=\frac{3}{4}$ so $x=2^\frac{3}{4}$.

Answer 2

There is a standard trick for such problems. If we're working over $\mathbb R$, remember that

$$x^{x^4}=8^2\iff \left(x^4\right)^{x^4}=8^8$$

Obviously $x^4=8=2^3$ is an only real solution. Thus,

$$\left(x^2-2^{\frac 32}\right)\left(x^2+2^{\frac 32}\right)=0$$

This implies the real solutions are

$$\color{gold}{\boxed{\color{black}{x_{1,2}=\pm 2^{3/4}}}}$$

On the other hand, we have

$$\ln x^4\times e^{\ln x^4}=8\ln 8\ge 0$$

and

$$\begin {aligned}&\ln x^4=W_0(24 \ln 2)\\ \implies &x^4=e^{W_0 (24\ln 2)}\end{aligned}$$

This gives us,

$$\color{gold}{\boxed{\color{black}{x_{1,2}=\pm e^{1/4W_0(24 \ln 2)}}}}$$

From here, you can also conclude that:

$$\color{gold}{\boxed{\color{black}{\pm 2^{3/4}=\pm e^{1/4W_0(24 \ln 2)}}}}$$