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If 3^(x + y) = 81 and 81^(x - y) = 9,then what is the value of the product xy? Express your answer as a common fraction.

 Jul 9, 2022
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Hi Guest!

Given: \(3^{(x+y)}=81=3^4\)

and: \(81^{(x-y)} = 9 \iff 3^{4(x-y)}=3^2\)

So, since the bases are the same, we get the following system of two equations:

\(x+y=4\)          (1)

\(4(x-y)=2\)     (2) 

\(\text{From (1): } x=4-y\), substitute in (2) to get:

\(4(4-y-y)=2 \implies 4(4-2y)=2 \implies (4-2y)=\dfrac{2}{4}=\dfrac{1}{2}\)

 

So: \(4-2y=\dfrac{1}{2} \implies y=\dfrac{4-\dfrac{1}{2}}{2}=\dfrac{7}{4}\)

 

So from (1): \(x=4-y=4-\dfrac{7}{4}=\dfrac{9}{4}\)

 

We want the product of x and y:

Therefore, \(xy=\dfrac{9}{4}*\dfrac{7}{4}=\dfrac{63}{16}\)

 

I hope this helps.

 Jul 9, 2022

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