# Suppose $x$ and $y$ are inversely proportional, if $x^p$ and $y^q$ are also inversely proportional, then how must $p$ and $q$ be related?

Suppose $$x$$ and $$y$$ are inversely proportional.

Then we have $$xy= k_1$$, $$x=\frac{k_1}{y}$$, $$y=\frac{k_1}{x}$$.

(a) If $$x^p$$ and $$y^q$$ are also inversely proportional, then how must $$p$$ and $$q$$ be related?

$$x^py^q= k_2$$
$$(\frac{k_1^p}{y^p})(\frac{k_1^q}{x^q})= x^py^q$$

(b) If $$x^p$$ and $$y^q$$ are directly proportional, then how must $$p$$ and $$q$$ be related?

$$\frac{x^p}{y^q}= k_2$$
$$(\frac{k_1^p/y^p}{k_1^q/x^q})= k_2 = k_1y^{p-q}(\frac{x^p}{y^p})$$

I'm completely lost on this one. I think I should be trying to get an equation in terms of $$p$$ and/or $$q$$ but I'm not sure how I can express this problem to arrive at a solution.

$$x^p (\frac{k_1^q}{x^q}) = k_2\\ x^{p-q} k_1^q = k_2\\ x^{p-q} = \frac{k_2}{k_1^q}$$
But the RHS is a constant that does not depend on $$x$$ or $$y$$. So the only way for this equation to be true is if the LHS is constant as well so $$p-q=0$$.
• $x^{p-q}=k_2k_1^q$ so $p-q$ is constant in both the situations described by (a) and (b). May 22, 2021 at 10:12