Test 2 Section 3 #7

This is one of my favorite questions because there are a couple fun paths to take: one with pure exponent rules, and one with exponent rules and difference of two squares. So fun! (I understand if you don’t share my enthusiasm.)

First, the exponent-rules-only method. We need to take what we’re given and simplify it enough that we can stop worrying about x and focus exclusively on a and b. To do that, we first need to remember that when we raise a power to a power, we multiply the powers. In other words:

    \begin{align*}\dfrac{x^{a^2}}{x^{b^2}}&=x^{16}\\\\\dfrac{x^{2a}}{x^{2b}}&=x^{16}\end{align*}

From there, we need to remember that when we divide exponential expressions with the same base, we subtract the exponents.

    \begin{align*}\dfrac{x^{2a}}{x^{2b}}&=x^{16}\\\\x^{2a-2b}&=x^{16}\end{align*}

From there, we’re good to eliminate x and just say that 2a-2b=16. Since the question asks us for the value of a-b, just divide that equation by 2 to get a-b=8.

Going the second way involves the division exponent rule but not the power-to-a-power rule. Skip right to the division rule:

    \begin{align*}\dfrac{x^{a^2}}{x^{b^2}}&=x^{16}\\\\x^{a^2-b^2}&=x^{16}\end{align*}

From there, we can say that a^2-b^2=16 and use our difference of two squares skills.

    \begin{align*}a^2-b^2&=16\\(a+b)(a-b)&=16\end{align*}

The question tells us that a+b=2, so let’s substitute:

    \begin{align*}(a+b)(a-b)&=16\\2(a-b)&=16\\a-b&=8\end{align*}

So there you go—two fun ways to get this one done. Which do you prefer?