Answer:
(x, y) = (0.5, 4) or (4, 0.5)
Explanation:
The given equations are more easily solved by making the bases of the logarithms all the same. Then we can use substitution to form a quadratic equation that will give two solutions to the problem.
Setup
Rewriting the first equation to use base-2 logarithms, we have ...
![\log_4(xy)=(1)/(2)\qquad\text{given}\\\\(\log_2(xy))/(\log_2(4))=(1)/(2)\qquad\text{change of base formula}\\\\(\log_2(x)+\log_2(y))/(2)=(1)/(2)\qquad\text{evaluate $\log_2(4)$, separate variables}\\\\\log_2(y)=1-\log_2(x)\qquad\text{solve for $\log_2(y)$}](https://img.qammunity.org/2023/formulas/mathematics/college/ztsc0woyjrnx8ur59yvkmgcu3ph6nugtot.png)
Solution
Substituting this expression into the second equation gives a quadratic in log₂(x).
![\log_2(x)(1-\log_2(x))=-2\qquad\text{substitute for $\log_2(y)$}\\\\\log_2(x)^2-\log_2(x)-2=0\qquad\text{put quadratic in standard form}\\\\(\log_2(x)-2)(\log_2(x)+1)=0\qquad\text{factor the quadratic}](https://img.qammunity.org/2023/formulas/mathematics/college/9ehoebovtcsmndsepssupbh4ndy63k3krs.png)
Solutions are values of log₂(x) that make the factors zero:
log₂(x) -2 = 0 ⇒ log₂(x) = 2 ⇒ x = 2² = 4
log₂(x) +1 = 0 ⇒ log₂(x) = -1 ⇒ x = 2⁻¹ = 1/2
The corresponding values of y are the other value of x:
log₂(y) = 1 -log₂(x) = 1 -2 = -1 ⇒ y = 1/2 for x = 4
log₂(y) = 1 -log₂(x) = 1 -(-1) = 2 ⇒ y = 4 for x = 1/2
Solutions are ...
(x, y) = (4, 1/2) or (1/2, 4)
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Additional comment
The attachment shows a graphing calculator solution to the original pair of equations.