# Given 2^(x-y)=6 and 2^(x+y)=12, what is y?

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### 1 Answer

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` .**