0
$\begingroup$

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.

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Only do one substitution so you get an equation in only one variable. This way you get to a simpler problem than with two variables. By doing both substitutions at the same time, you were getting at the same level.

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

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 $p-q=0$.

$\endgroup$
2
  • $\begingroup$ What exactly do you mean by "Getting at the same level"? Is that to say we can keep recursively substituting without really changing the result? $\endgroup$
    – Oxidizing2408
    May 22, 2021 at 10:10
  • $\begingroup$ $x^{p-q}=k_2k_1^q$ so $p-q$ is constant in both the situations described by (a) and (b). $\endgroup$
    – Oxidizing2408
    May 22, 2021 at 10:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .