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Given 2^(x-y)=6 and 2^(x+y)=12, what is y?
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You need to use exponentiation rules, such that:
`2^(x - y) = 2^x/2^y`
`2^(x+y) = 2^x*2^y`
The problem provides the information that `2^(x - y) = 6` , hence, replacing `2^x/2^y` for `2^(x - y)` yields:
`2^x/2^y = 6`
The problem also provides the information that `2^(x + y) = 12` , hence, replacing `2^x*2^y` for `2^(x + y)` yields:
`2^x*2^y = 12 => 2^x*2^y = 6*2`
Replacing `2^x/2^y` for 6 yields:
`2^x*2^y = 2^x/2^y*2`
Reducing duplicate factors yields:
`2^y = 2/2^y => 2^y = 2^(1 - y)`
Equating the exponents, yields:
`y = 1 - y => 2y = 1 => y = 1/2`
Hence, evaluating y, under the given conditions and using the exponentiation rules, yields `y = 1/2` .
Posted by sciencesolve on August 9, 2013 at 3:54 PM (Answer #1)
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