SOLUTION: If X^log₄Y + Y^log₄X = 4 and log₄X

SOLUTION: If X^log₄Y + Y^log₄X = 4 and log₄X - log₄Y = 1, find X and Y.

Question 1209767: If X^log₄Y + Y^log₄X = 4
and log₄X - log₄Y = 1,
find X and Y.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's solve this system of equations step-by-step.
1. Simplify the Second Equation:
log₄X - log₄Y = 1
log₄(X/Y) = 1
X/Y = 4¹
X = 4Y
2. Substitute X = 4Y into the First Equation:
X^(log₄Y) + Y^(log₄X) = 4
(4Y)^(log₄Y) + Y^(log₄(4Y)) = 4
3. Simplify the Terms:
(4Y)^(log₄Y) = 4^(log₄Y) * Y^(log₄Y) = Y * Y^(log₄Y) = Y^(1 + log₄Y)
Y^(log₄(4Y)) = Y^(log₄4 + log₄Y) = Y^(1 + log₄Y)
4. Substitute the Simplified Terms Back into the Equation:
Y^(1 + log₄Y) + Y^(1 + log₄Y) = 4
2Y^(1 + log₄Y) = 4
Y^(1 + log₄Y) = 2
5. Take log₄ of Both Sides:
log₄(Y^(1 + log₄Y)) = log₄2
(1 + log₄Y)log₄Y = 1/2
6. Let u = log₄Y:
(1 + u)u = 1/2
u² + u = 1/2
u² + u - 1/2 = 0
7. Solve the Quadratic Equation for u:
Multiply by 2 to get rid of the fraction:
2u² + 2u - 1 = 0
Use the quadratic formula:
u = [-b ± √(b² - 4ac)] / 2a
u = [-2 ± √(2² - 4(2)(-1))] / (2 * 2)
u = [-2 ± √(4 + 8)] / 4
u = [-2 ± √12] / 4
u = [-2 ± 2√3] / 4
u = [-1 ± √3] / 2
Since Y must be positive, log₄Y must be real, so we keep both solutions for u.
8. Solve for Y:
log₄Y = u
Y = 4^u
Case 1: u = (-1 + √3) / 2
Y = 4^((-1 + √3) / 2)
Case 2: u = (-1 - √3) / 2
Y = 4^((-1 - √3) / 2)
9. Solve for X:
X = 4Y
Case 1:
X = 4 * 4^((-1 + √3) / 2) = 4^((1 - 1 + √3) / 2) = 4^(√3 / 2)
Case 2:
X = 4 * 4^((-1 - √3) / 2) = 4^((1 - 1 - √3) / 2) = 4^(-√3 / 2)
Solutions:
Solution 1: X = 4^(√3 / 2), Y = 4^((-1 + √3) / 2)
Solution 2: X = 4^(-√3 / 2), Y = 4^((-1 - √3) / 2)
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