Finding the value of $x^{x^2}+x^{x^6}$ given that $x^{x^4}=4$

Question

If $x^{x^4}=4$; Then what is the value of $x^{x^2}+x^{x^6}$ ; I just want a hint like from where should I start. (other than using a trial and error to find that $x=\sqrt{2}$ satisfies this and now just put it in the required equation!!)

Answer 1

You only ask for a hint, so here is one:

In $x^{x^4}=4$, raise both sides to the power $4$. This will give you $$ (x^4)^{x^4}=4^4. $$ It is clear that $x^4=4$ is one possibility, but maybe you can conclude that it is the only one, by studying the function $$ c\mapsto c^c? $$

Edit Let us answer the following: What positive values of $c$ give $c^c=4^4$?

Let $f(c)=c^c$. Then $$ f'(c)=c^c(1+\ln c), $$ so, $f$ is monotonically decreasing for $0<c<1/e$ and monotonically increasing for $c>1/e$. Since $\lim_{c\to 0^+}f(c)=1$ and $f(1)=1$, we conclude that the equation $c^c=4^4$ is only attained for one $c$, namely the "obvious" $c=4$.

If we go back to your problem, this means that (here $c=x^4$) $$ (x^4)^{x^4}=4^4\quad\implies\quad x^4=4. $$ Ask for more information, if needed.