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Re: If x and y are integers and 12^x∗6^y=432, what is the value of xy? [#permalink]

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21 Apr 2017, 23:15

option:C Time taken:1:53(Should be less than 1:30) 12^x*6^y=432 (2^2*3^1)^x*(2*3)^y=2^4*3^3 2^2x*3^x*2^y*3^y=2^4*3^3 Base same power add,So 2^2x+y*3^x*y=2^4*3^3 So we get 2x+y=4--(a) and x+y=3--(b) Comparing and subtracting a from b x=1-- put value in b 1+y=3 y-2 so xy=(1)(2)=2
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Re: If x and y are integers and 12^x∗6^y=432, what is the value of xy? [#permalink]

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22 Apr 2017, 23:00

Bunuel wrote:

If x and y are integers and 12^x∗6^y=432, what is the value of xy?

A. 0 B. 1 C. 2 D. 3 E. 4

In order to solve this question efficiently we should break down 432

[12^x] x [6^y]= 432 = 6 x 72 = 6 x 8 x 9 = 6 x 2^2 x 3^2 = 6 x 6 x 2^2 x 3 = 6^2 2^2 3 = 6^2 x 6 x 2 = 6^2 x 12^1 (12 instead of 6^3 so you can the same bases as the other side of the equation- also it is important to know it is 12 to first power still) [12^x] x [6^y] = 6 ^2 x 12^1 xy = (2)(1) (because we have similar bases- this is a concept important in calculus also- antiderivatives)

While we cannot solve an equation with variables we can solve the product, which is what question asks us to do, by forming the same bases on the left of the equation through the factorization of 432.

If x and y are integers and 12^x∗6^y=432, what is the value of xy?

A. 0 B. 1 C. 2 D. 3 E. 4

We can simplify the equation:

12^x ∗ 6^y = 432

(2^2 * 3)^x * (2 * 3)^y = 8 * 54

2^2x * 3^x * 2^y * 3^y = 2^3 * 2 * 3^3

2^2x * 2^y * 3^x * 3^y = 2^3 * 2 * 3^3

2^(2x + y) * 3^(x + y) = 2^4 * 3^3

We can equate the exponents of each variable. Thus, 2x + y = 4 and x + y = 3. Subtracting the two equations, we have x = 1. Furthermore, we have 1 + y = 3, or y = 2. Thus, the value of xy = 1(2) = 2.

Answer: C
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