Solve for x and y in the logarithmic system of equations:

Question

8 (√2)^x-y = .5^y-3
log{3}(x-2y)+ log{3}(3x+2y) = 3

Answer 1

The second equation can be simplified: log(x-2y)(3x+2y)=3 (log to base 3). Therefore, raising both sides as powers 0f 3, (x-2y)(3x+2y)=3^3=27. So 3x^2-4xy-4y^2-27=0.

The first equation can be written: (2^3)2^((x-y)/2)=(1/2)^(y-3)=2^(3-y)=(2^3)(2^(-y)).

(2^3) cancels out on each side leaving: 2^((x-y)/2)=2^(-y). The exponents can be equated: (x-y)/2=-y or x-y=-2y so x=-y. We can now substitute x=-y or y=-x into the other simplified equation: 3x^2+4x^2-4x^2-27=0. 3x^2=27 and x^2=9, making x=3 or -3.