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ggpapas
if 2^(x+y) + 3^(x-y) = 32/3 then what is the value of product x and y?

A.2
B.3
C.5
D.6
E.16

Hi ggpapas,

Can you please check if the question is correct. I tried for almost 45 minutes to figure out the answer but couldn't get any approach.
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ggpapas
if 2^(x+y) + 3^(x-y) = 32/3 then what is the value of product x and y?

A.2
B.3
C.5
D.6
E.16

It seems that there is a typo and it should be multiplication instead of addition. Please check once again.
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Hi all,

Yes I had a typo, the addition should be multiply

Thanks!
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ggpapas
if 2^(x+y) x 3^(x-y) = 32/3 then what is the value of product x and y?

A.2
B.3
C.5
D.6
E.16

\(2^{x+y}*3^{x-y}=\frac{32}{3}=\frac{2^5}{3}=2^5*3^{-1}\)..

So by equating powers
x+y=5 ...
x-y=-1.....x=y-1
So x+y=y-1+y=5.......2y=6....y=3
x=y-1=3-1=2..
xy=3*2=6

D
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ggpapas
if \(2^{(x+y)} * 3^{(x-y)} = \frac{32}{3}\) then what is the value of product x and y?

A.2
B.3
C.5
D.6
E.16


Theory to solve such questions
The key to solving these kind of problems is to equate the left hand side to the
right hand side. Once you have a common base on both the side the exponents
can be equated.

For an equation \(Base1^{Exponent1} = Base2^{Exponent2}\)

If \(Base1 = Base2\), then \(Exponent1 = Exponent 2\)


Now, coming back to this question.

\(2^{(x+y)} * 3^{(x-y)} = \frac{32}{3}\) can be re-written as \(2^{(x+y)} * 3^{(x-y)} = 2^5 * 3^{-1}\)

Using this equation we get x+y = 5 and x-y = -1
Solving for x and y, we get x=2 and y=3
The product of x and y is 6(Option D)
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ggpapas
If \(2^{(x+y)} * 3^{(x-y)} = \frac{32}{3}\) then what is the value of product x and y?

A. 2
B. 3
C. 5
D. 6
E. 16
32/3 = 2^5 * 3^ (-1)
x+y = 5
x-y = -1
2x = 4
x= 2, y =3
xy = 6
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