anyone can help me with solving this $x^{x^{3}}=3$? Find the value of $x$ : $x^{x^{3}}=3$
I tied with "log" but I couldn't. any help?
 A: A nice first step toward the solution is to increase the symmetry of the LHS: if you cube $x^{x^3}$, you get $x^{3x^3}$, which is $(x^3)^{x^3}$.  Symmetry!  Since cubing the RHS gives $3^3$, it's easy to then spot the solution $x^3=3$.
A: consider $$x^{x^b}=a :x>0$$
take log for both side:
$$x^b\log x=\log a$$
let $e^t=x$
$$e^{bt}t=\log a$$
$$e^{at}bt=b\log a$$
so
$$bt=W(b\log a)$$
$$t=\frac{W(b\log a)}{b}$$
$$x=e^{\frac{W(b\log a)}{b}}$$
where $W(R)$ is Lambert W function
for $a=b=3$
$x \approx 1.44225 $
another try:
$${\color{Red} c}=x^c \Rightarrow x=\sqrt[c]{c}$$
$$\Rightarrow x^{{\color{Red} {x^c}}}=c$$
$$ \Rightarrow x=\sqrt[c]{c}$$
A: Assume the domain is $(0,\infty)$. You show the equation $x^{x^3} = 3$ has a unique real solution $x = \sqrt[3]{3}$. Look at the function:  $f(x) = x^3\ln x , 0 < x < \infty$. If $0 < x < 1 \Rightarrow x^3\ln x < 0$, so $x^3\ln x < \ln 3$, since $\ln 3 > 0$. Thus: $e^{x^3\ln x} < e^{\ln 3}\Rightarrow x^{x^3} < 3$. And for $ 1\leq x <\infty$, the function $f(x) = x^3\ln x$ has $f'(x) = 3x^2\ln x+ x^2 = x^2(3\ln x + 1) > 0$. This means the equation $x^{x^3} = 3$ can only have atmost $1$ solution on $[1,\infty)$. Observe that $x = \sqrt[3]{3}$ is a solution, and so it is the only solution.
A: Hint: $\large (\sqrt[3]{3})^3=3{}{}$.
