Exponential Equations with Constraints: Finding the Sum of Two Exponents

[yoleven]

Homework Statement

for x and y satisfying x-y=2 and 2^{x}-2^{y}=6
find 2^{x}+2^{y}.

The Attempt at a Solution

x-y=2 ; x=2+y
log_{2} 2^{x}-log_{2}2^{y}=log_{2}6

log 6/log2=2.585

x-y=2.585 but this violates the intitial condition x-y=2

where am i going wrong?

 

[nietzsche]

I don't think you need to use logs.

x-y=2
x=y+2

2^(y+2) - 2^y = 6

I think you should be able to figure it out from that.

 

[Bohrok]

Yes, use substitution. Also
log₂(a + b) \neq log₂a + log₂b

 

[yoleven]

I'm not seeing what to do. I see the substitution but then I don't know what to do.

 

[nietzsche]

Let's take
x - y = 2
x = y + 2

substitute into your equation

2^y+2 - 2^y = 6
(2^y)(2²) - 2^y = 6
2^y(4-1) = 6
...

You don't even have to solve for x and y explicitly.

 

[yoleven]

thank you very much.