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xy=32 2xy^2=32 find the value of p and q when x=2^p and y=2^q
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@satellite73
\[xy=32\\ 2xy^2=32\] right?
s
ok so we can solve for \(x\) and \(y\) by dividing
\[\frac{xy}{2xy^2}=\frac{32}{32}=1\] or \[\frac{1}{2y}=1\]
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that makes \(y=\frac{1}{2}\) substitute this back in to either equation to find \(x\)
you good from there?
x = 64
yes
\[64=2^6,\frac{1}{2}=2^{-1}\]
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