I’m practicing index laws and I keep mixing up when to add exponents and when to multiply them, especially once negatives and parentheses jump in. The expression I’m trying to simplify is:
(2x^-3 y^2)^-2 ÷ (4x y^-1)^-1
I know the rules in theory: same base multiply -> add exponents, power of a power -> multiply exponents, division -> subtract exponents, negative exponent -> reciprocal. But when I try to apply them all at once, I get tangled.
My attempt:
– For the first part, (2x^-3 y^2)^-2, I wrote: 2^-2 x^(+6) y^-4. I think (x^-3)^-2 gives x^6, and (y^2)^-2 gives y^-4. The 2 becomes 2^-2 (right?).
– For the second part, (4x y^-1)^-1, I wrote: 4^-1 x^-1 y^1. I figured the -1 on y^-1 flips it back to y^1.
– Then I tried to divide: [2^-2 x^6 y^-4] ÷ [4^-1 x^-1 y^1]. For the variables, I did x: 6 – (-1) = 7, and y: -4 – 1 = -5. But for the numbers I froze: is 2^-2 ÷ 4^-1 the same as 2^-2 * 4^1? Or should I be combining them some other way?
Could someone show a clean, step-by-step way to handle this without skipping steps, and point out exactly where my reasoning goes off? I mainly get confused about the coefficients with negative exponents during division, and keeping the signs straight.
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3 Responses
Distribute the outer powers first: (2x^-3 y^2)^-2 = 2^-2 x^6 y^-4 and (4x y^-1)^-1 = 4^-1 x^-1 y, then divide by subtracting exponents and treating the numbers like ordinary fractions-2^-2 ÷ 4^-1 = (1/4) ÷ (1/4) = 1, so x: 6 − (−1) = 7 and y: −4 − 1 = −5-giving x^7 y^-5 = x^7/y^5.
When I was learning this I literally wrote “outside multiplies, across divide subtract; a negative flips levels” on a sticky note, and once I treated 2^-2 and 4^-1 as 1/4 ÷ 1/4, the confusion vanished.
When exponents start throwing confetti, I like to tame them by distributing the outside power first. So (2x^-3 y^2)^-2 becomes 2^-2 x^6 y^-4 (power of a power multiplies the exponents, signs included). And (4x y^-1)^-1 becomes 4^-1 x^-1 y^1. Now divide term-by-term: for x, 6 − (−1) = 7; for y, −4 − 1 = −5. The coefficient bit is where gremlins hide, but it’s actually neat: 2^-2 ÷ 4^-1 = 2^-2 ÷ (2^2)^-1 = 2^-2 ÷ 2^-2 = 2^0 = 1 (equivalently, 1/4 ÷ 1/4 = 1), so the numbers cancel. That leaves x^7 y^-5, which is x^7 / y^5 with positive exponents. If signs feel slippery, an alternative is to flip negatives first: (2x^-3 y^2)^-2 = 1/(4 x^-6 y^4) = x^6/(4y^4) and (4x y^-1)^-1 = y/(4x); then (x^6/(4y^4)) ÷ (y/(4x)) = x^7/y^5. Your variable exponents were handled perfectly-the only snag was recognizing 4^-1 = 2^-2, so the coefficients vanish. Which method feels friendlier to you: distribute the power first, or turn all the negatives into reciprocals before dividing?
I go “power to each factor” first: (2x^-3 y^2)^-2 = 2^-2 x^6 y^-4 and (4x y^-1)^-1 = 4^-1 x^-1 y, then divide base-by-base-x: 6 − (−1) = 7, y: −4 − 1 = −5, and for the numbers 2^-2 ÷ 4^-1 = (1/4) ÷ (1/4) = 1-so the result is x^7 y^-5 = x^7 / y^5.
Nice refresher on why these rules behave this way: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:exp-properties.