Rewriting expression involving power functions Can you please help me with the following equation from a pre-calculus course? As far as I know we can perform the division when the denominator and numerator have the same base (but could have different powers). But I see no way of achieving it here.
For which a and b does the following hold (assume x>0 and y>0):
$${\sqrt{\frac{(x^2y)^{-3}}{y^{-1}x}}} = x^ay^b$$
 A: There are several properties which you can use here:
$1$) $(rs)^t=r^ts^t$
Therefore $(x^2y)^{-3}=(x^2)^{-3}y^{-3}$
$2$) $(r^s)^t=r^{st}$
Therefore $(x^2)^{-3}y^{-3}=x^{-6}y^{-3}$
$3$) $st=ts$
Therefore $y^{-1}x=xy^{-1}$
$4$) $\dfrac{st}{uv}=\dfrac{s}{u}\cdot\dfrac{t}{v}$
Therefore $\dfrac{x^{-6}y^{-3}}{xy^{-1}}=\dfrac{x^{-6}}{x}\cdot\dfrac{y^{-3}}{y^{-1}}$
$5$) $\dfrac{x^m}{x^n}=x^{m-n}$
Therefore $\dfrac{x^{-6}}{x}\cdot\dfrac{y^{-3}}{y^{-1}}=x^{-7}y^{-2}$
$6$) If $s>0$ and $t>0$ then $\sqrt{st}=\sqrt{s}\sqrt{t}$
Therefore $\sqrt{x^{-7}y^{-2}}=\sqrt{x^{-7}}\sqrt{y^{-2}}$
$7$) If $s>0$ then $\sqrt{s^r}=s^{r/2}$
Therefore, $\sqrt{x^{-7}}\sqrt{y^{-2}}=x^{-7/2}y^{-1}$
So $a=-\dfrac{7}{2}$ and $b=-1$.
A: $$\begin{align}
{\sqrt{\frac{(x^2y)^{-3}}{y^{-1}x}}} &= x^ay^b \\[3pt]
\frac{(x^2y)^{-3}}{y^{-1}x} &= x^{2a}y^{2b} \tag{$x,y>0$}\\[3pt]
(x^2y)^{-3} &= x^{2a+1}y^{2b-1} \\[3pt]
x^{-6}y^{-3} &= x^{2a+1}y^{2b-1} \\[3pt]
1 &= x^{2a+7}y^{2b+2} \\
\end{align}$$
If you need this to hold no matter what the values of $x>0$ and $y>0$ are, then
$$\begin{align}
2a+7 &= 0 \\[3pt]
a &= -\frac 72 \\[3pt]
2b+2 &=0 \\[3pt]
b &= -1 \\
\end{align}$$
