$x$ such that $\sqrt{x^y} > \sqrt{(x y)^2}$ Let $y>1$, $x>0$
Of what form are numbers $x$ such that:
$$\sqrt{x^y} > \sqrt{(x y)^2}$$
I.e: What is the solution for $x$?
EDIT:
As the first answer implied:
$$\sqrt{x^y} > \sqrt{(x y)^2} \implies \log(\sqrt{x^y}) > \log(\sqrt{(x y)^2})\implies$$
$$\log(x^y) > \log((x y)^2) \implies y \log(x)>2 \log(x y) \implies y \log(x)> (\log(x)+\log(y))$$
$$\implies (y-2) \log(x)>2 \log(y)$$
 A: To properly make sense of $\sqrt{x^y}$, let us assume that $x \ge 0$. Using that $\log, \exp$ and $\sqrt\cdot$ are strictly increasing on their domain, we obtain:
\begin{align}
\sqrt{x^y} > \sqrt{(xy)^2} &\iff x^y > (xy)^2 \\
&\iff y \log x > 2(\log x +\log y)\\
&\iff (y-2)\log x > 2 \log y\\
&\iff \log x > \frac{2 \log y}{y-2} \quad\text{if $y \ne 2$}\\
&\iff x > y^{2/(y-2)}
\end{align}
Because the inequality at step $3$ does not hold for $y = 2$, we may discard this possibility from our final solution.
A: obviously x has to be positive. now take logarithms of both sides, the inequality you get is equivalent, so you have $1/2y \log(x) > \log(x) + \log(y)$, now solve for $\log(x)$ and you're done
oh, sorry, actually if you allow y to be an even natural number then the inequality still makes sense even for negative x's, but it's still pretty much the same, because then it's the same for x as for -x so doing it by my method you can find solutions with assumption $x > 0$. then you just take the set of these solutions and you also add all their negatives - so in this case x is a solution iff -x is a solution
you might want to clarify the question tho' because in general we don't define $x^y$ for $x <0$ it's more of a philosophical question
