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
Many students will first try to use the given information to first determine the values of x and y, and THEN use those values to evaluate the expression at hand.
However, we can save some time by recognizing that....
(2^x)/(
2^1) = 2^(x-1)
and (3^y)/(
3^2) = 3^(y-2)
GIVEN: (2^x)(3^y) = 288
Divide both sides by
2^1 (aka
2) to get: (2^x)(3^y)/(
2^1) = 288/
2 Simplify both sides to get: [2^(x-1)](3^y) = 144
Divide both sides by
3^2 (aka
9) to get: [2^(x-1)](3^y)/(
3^2) = 144/
9Simplify both sides to get: [2^(x-1)][3^(y-2)] = 16
Answer: 16
Cheers,
Brent
_________________
Brent Hanneson - founder of Greenlight Test Prep