SOLUTION: If 2^x + 2^y = 6 and x + y = 2, find 4^x + 4^y.

Question 1209743: If 2^x + 2^y = 6
and x + y = 2,
find 4^x + 4^y.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to solve this problem:
**1. Express y in terms of x:**
Since x + y = 2, we can write y = 2 - x.
**2. Substitute y in the first equation:**
2^x + 2^(2-x) = 6
**3. Simplify the equation:**
2^x + (2^2 / 2^x) = 6
2^x + (4 / 2^x) = 6
**4. Introduce a substitution:**
Let u = 2^x. Then the equation becomes:
u + (4/u) = 6
**5. Solve for u:**
Multiply the entire equation by u:
u^2 + 4 = 6u
u^2 - 6u + 4 = 0
**6. Use the quadratic formula to solve for u:**
u = (-b ± sqrt(b^2 - 4ac)) / 2a
u = (6 ± sqrt((-6)^2 - 4 * 1 * 4)) / 2 * 1
u = (6 ± sqrt(36 - 16)) / 2
u = (6 ± sqrt(20)) / 2
u = (6 ± 2√5) / 2
u = 3 ± √5
**7. Find the two possible values for 2^x:**
So, 2^x = 3 + √5 or 2^x = 3 - √5
**8. Find the corresponding values for 2^y:**
Since 2^x * 2^y = 2^(x+y) = 2^2 = 4:
If 2^x = 3 + √5, then 2^y = 4 / (3 + √5) = 4(3 - √5) / ((3 + √5)(3 - √5)) = 4(3 - √5) / (9 - 5) = 3 - √5
If 2^x = 3 - √5, then 2^y = 4 / (3 - √5) = 4(3 + √5) / ((3 - √5)(3 + √5)) = 4(3 + √5) / (9 - 5) = 3 + √5
Notice that the values of 2^x and 2^y are just swapped.
**9. Calculate 4^x + 4^y:**
4^x + 4^y = (2^x)^2 + (2^y)^2
Case 1: 2^x = 3 + √5 and 2^y = 3 - √5
4^x + 4^y = (3 + √5)^2 + (3 - √5)^2 = (9 + 6√5 + 5) + (9 - 6√5 + 5) = 14 + 14 = 28
Case 2: 2^x = 3 - √5 and 2^y = 3 + √5
4^x + 4^y = (3 - √5)^2 + (3 + √5)^2 = (9 - 6√5 + 5) + (9 + 6√5 + 5) = 14 + 14 = 28
In either case, 4^x + 4^y = 28.
**Final Answer:** 4^x + 4^y = 28

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