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?

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.
