Stm1. 2^3*3^2=72CappyAA wrote:If x and y are positive integers, what is the value of x+y?
(1) (2^x)(3^y) = 72
(2) (2^x)(2^y) = 32
Thanks,
Cappy
x+y=5 Sufficient
Stm2. 2^2*2^3=32
x+y=5 Sufficient
D
What is the value of x+y?
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foobarnull
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Thanks for this. I get the exponents completely but I'm... wondering what's the arithmetic work behind this? I wasn't aware that there's an operation for multiplying different exponents on different bases. Please help... thanks.aspire750 wrote: Stm1. 2^3*3^2=72
x+y=5 Sufficient
Stm2. 2^2*2^3=32
x+y=5 Sufficient
D
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Ian Stewart
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Ah, no- you're right, and there is no rule for 'multiplying exponents on different bases'. aspire750 was just using prime factorizations to solve the problem. If you look, for example, at statement 1), we have:foobarnull wrote:Thanks for this. I get the exponents completely but I'm... wondering what's the arithmetic work behind this? I wasn't aware that there's an operation for multiplying different exponents on different bases. Please help... thanks.aspire750 wrote: Stm1. 2^3*3^2=72
x+y=5 Sufficient
Stm2. 2^2*2^3=32
x+y=5 Sufficient
D
(2^x)(3^y) = 72
We know x and y are positive integers. Prime factorize 72:
72 = (2^3)*(3^2)
so we know
(2^x)(3^y) = (2^3)*(3^2)
Since x and y are positive integers, the left side and the right side of this equation are both prime factorizations. By the aptly named 'Fundamental Theorem of Arithmetic', prime factorizations are unique. That is, if one prime factorization equals another, the powers on each prime must be the same. If
(2^x)(3^y) = (2^3)*(3^2)
then the power on 2 on the left side must be equal to the power on 2 on the right side: x must be equal to 3. Same with the power on 3: y must be 2. Since we know x and y, we can add them: x+y = 5.
The second statement is a bit easier (though I'd note that, using Statement 2 alone, you cannot conclude as aspire750 did that x = 3 and y = 2; there are other possibilities):
(2^x)(2^y) = 32
2^(x+y) = 2^5
x+y = 5
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