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3 Responses
You’ve got it: x^5/x^{-2} = x^{7} and (2y^{-3})^{0} = 1-negative exponents mean “take the reciprocal,” so a^{-4} = 1/a^{4}, and dividing by 1/a^{4} is the same as multiplying by a^{4} (like flipping a fraction to undo a divide); just avoid y = 0 and remember 0^0 is undefined.
Example: 6 ÷ 2^{-3} = 6 × 2^{3} = 48.
I feel you-exponents can look like they’re doing gymnastics! The quick checks: yes, x^5/x^-2 = x^(5−(−2)) = x^7, and (2y^-3)^0 = 1 as long as y ≠ 0 (any nonzero base to the 0 is 1). Intuition: a negative exponent just means “reciprocal,” so a^-4 = 1/a^4; if that’s in the denominator, you’re dividing by 1/a^4, which flips to multiplying by a^4-like dividing by a tiny fraction makes things bigger. A simple example: 2^5/2^-2 = 32 ÷ (1/4) = 128 = 2^7. Caveats to keep you out of trouble: don’t let the base be zero when negatives are around (a^-4 needs a ≠ 0), and 0^0 is undefined; also, parentheses matter-2y^-3 means 2/y^3, but (2y)^-3 means 1/(8y^3). And remember, these subtraction-and-flip rules only combine exponents when the bases match. You’ve got the right idea-keep treating negatives as “reciprocals” and the flips will start to feel natural.
Yep, your instincts are spot on: x^5/x^-2 = x^(5-(-2)) = x^7, and (so long as y ≠ 0) (2y^-3)^0 = 1-I remember the lightbulb moment when I wrote a^-4 = 1/a^4 and then 1/(a^-4) = a^4, which made the “flip” feel like a simple reciprocal move rather than magic.
Caveats: watch for zero (y can’t be 0, 0^-n is undefined, and 0^0 is indeterminate), and here’s a crisp refresher from Khan Academy: https://www.khanacademy.org/math/algebra/exponent-equations/exponents-alg1/a/negative-exponents.