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

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Brent Hanneson – Creator of greenlighttestprep.com

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