SOLUTION: Solve the system. Log base x (2y) =3 Log base x (4y)=2 I did the change of x^3 = 2y then 2y = x^3 for the first equation. Then, I did the same for the second one. x^2 = 4y.

Algebra ->  Systems-of-equations -> SOLUTION: Solve the system. Log base x (2y) =3 Log base x (4y)=2 I did the change of x^3 = 2y then 2y = x^3 for the first equation. Then, I did the same for the second one. x^2 = 4y.      Log On


   

Question 373868: Solve the system.
Log base x (2y) =3
Log base x (4y)=2
I did the change of x^3 = 2y then 2y = x^3 for the first equation. Then, I did the same for the second one. x^2 = 4y. I was trying to solve for y on the first one but I am confused now.
Thank you.
EC
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
2y+=+x%5E3
y+=+%281%2F2%29%2Ax%5E3
and
4y+=+x%5E2
y+=+%281%2F4%29%2Ax%5E2
I can say
%281%2F2%29%2Ax%5E3+=+%281%2F4%29%2Ax%5E2
Multiply both sides by 4
2x%5E3+=+x%5E2
Divide both sides by x
2x+=+1
x+=+1%2F2
and, since
2y+=+x%5E3
2y+=+1%2F8
y+=+1%2F16
and, also,
4y+=+x%5E2
4y+=+1%2F4
y+=+1%2F16
OK