amorphous wrote:
if \(2^x * 3^y = 288\), where x and y are +ve integers then \((2^{x-1})(3^{y-2})\)is equal to?
src:orbit test prep
APPROACH #1: Use prime factorization\(288 = (2)(2)(2)(2)(2)(3)(3) = (2^5)(3^2)\)
So, the given equation becomes: \((2^x)(3^y) = (2^5)(3^2)\), which means \(x=5\) and \(y=2\)
Thus \((2^{x-1})(3^{y-2}) = (2^{5-1})(3^{2-2}) = (2^{4})(3^{0}) = (16)(1) = 16\)
Answer: 16APPROACH #2: Use algebraic manipulationGiven: \((2^x)(3^y) = 288\)
Divide both sides of the equation by \(2^1\) (aka \(2\)) to get: \(\frac{(2^x)(3^y)}{2^1} = \frac{288}{2^1}\)
Simplify to get: \((2^{x-1})(3^y) = 144\)
Divide both sides of the equation by \(3^2\) to get: \(\frac{(2^{x-1})(3^y)}{3^2} = \frac{144}{3^2}\)
Simplify: \((2^{x-1})(3^{y-2}) = 16\)
Answer: 16
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Brent Hanneson - founder of Greenlight Test Prep