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.
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.
