What is the most efficient strategy to symplify : $\left(\frac{2u^-5v^2}{8w}\right)^{-2}$ at minimal coast, I mean, with the minimal amount of necessary and sufficient manipulations?
A slightly different question : how would you actually proceed during a test such as Barton's?
Let's say that every application of an algebra rule or of an arithmetic fact counts here as a manipulation.
Source : Barton College Placement Test ( https://www.barton.edu/pdf/math/practice-math-placement-test.pdf)
I can't do better than at least 9 manipulations :
$\left(\frac{2u^{-5}v^2}{8w}\right)^{-2}$
= $\left(\frac{1u^{-5}v^2}{4w}\right)^{-2}$ ( by 2/8 = 1/4)
= $\left(\frac{1u^{-5}v^2}{4w}\right)^{(-1)(2)}$ ( by $-2$ = $-1\times 2$)
= $\left(\frac{1u^{-5}v^2}{4w}\right)^{(2)(-1)}$ ( by : mult. commutativity )
= $\left(\frac{1^2u^{-10}v^4}{16w^2}\right)^{(-1)}$ ( by $\left(\frac ab\right)^n$= $\frac{a^n}{b^n}$ , by $(ab)^n = a^n\times{b^n}$ and by $(a^n)^m$ = $a^{nm}$, applied several times).
= $\left(\frac{16w^2}{1^2u^{-10}v^4}\right)$ ( by $\left(\frac ab\right)^{-1}$ = $\frac{b}{a}$ )
= $\left(\frac{16w^2}{1^2v^4}\frac{1}{u^{-10}}\right)$ ( by : 1 is the $\times$ identity and by $\times$ commutativity)
= $\left(\frac{16w^2u^{-(-10)}}{1^2v^4}\right)$ ( by $a^{-n}= \frac{1}{a^n}$)
= $\left(\frac{16w^2u^{10}}{1^2v^4}\right)$ ( by : $ - (-a) = a$)
= $\left(\frac{16w^2u^{10}}{v^4}\right)$ ( by $1^2 = 1$ and $1$ is the identity element for $\times$).
Fix all negative exponents, starting with the outer-most. $(a/b)^{-n} = (b/a)^n$
$$\left(\frac{2u^{-5}v^2}{8w}\right)^{-2} =\left(\frac{8w}{2u^{-5}v^2}\right)^{2} $$
$$ = \left(\frac{8wu^5}{2v^2}\right)^{2} $$
Reduce
$$\left(\frac{4wu^5}{v^2}\right)^{2}$$
Distribute exponent
$$ = \frac{(4wu^5)^2}{(v^2)^2} $$
Square each
$$ = \frac{16w^2u^{10}}{v^4} $$