# Exponential Sytems of Equations

1. Apr 14, 2015

### ConstantineO

1. The problem statement, all variables and given/known data
Solve the following system.

2. Relevant equations

(2^(x+y))=16, (2^(x-y)) = 4

3. 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.

2. Apr 14, 2015

### Staff: Mentor

The easiest way to do this problem is to multiply the two equations together (left sides time left sides, and right sides time right sides). This eliminates y and immediately gives you x. Another easy way to do this problem is to take the logs of both equations to the base 2.

Chet

3. Apr 15, 2015

### ConstantineO

I feel like such a fool sometimes... It's this stuff right here that makes me question if I can get into university. The answer and methods of finding the solution were so simple, but I just couldn't see it. Thank you very much.

4. Apr 15, 2015

### Staff: Mentor

Please don't get discouraged. We were taught to think in different ways like this when we were on math team in HS. You just need to spend more time thinking about other options for how to solve the problem before you really dive in.

Chet