- #1
ConstantineO
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Homework Statement
Solve the following system.
2. Homework Equations
(2^(x+y))=16, (2^(x-y)) = 4
The Attempt at a Solution
(2^(x+y))=16
(2^x)(2^y) = 16
EQ1 [(z)(q) = 16]
2^(x-y)) = 4
(2^x)/(2y) = 4
EQ2 [(z)/(q)=4]
Let z = (2^x)
Let q = (2^y)
Isolate the "z" variable in EQ1 so it can be subbed into EQ2
(z)(q) = 16
(z) = (16)/(q)
EQ2 Substitution Step
(z)/(q)=4
((16)/(q))/(q) = 4
q = (4)((16)/(q))
q = (64)/(q)
q^2 = 64
q = 8
Since we know "q" = 8, we can use this to discover the value of the "y" variable in (2^y)
q = 8
(2^y) = 8
y =log2(8)
y= 3
Now this is where I am lost and somewhat confused. If I sub the value for " y" into equation "(2^(x+y))=16," I find the x variable to be "1". This introduces a problem with the second equation:
If y = 3 and x = 1
2^(x-y)) = 4
2^(1-3)) = 4
2^(-2) = 4
1/(2^2) =4
1/4 =/= 4
I know what the answers are and how they must be arranged, but how do I find the "1" in the most elegant and correct way possible?
EDIT: I know how to spell systems. I made a typo. Please don't crucify me.