Logarithmic system of equations Solve these equations log x+(log x+8 log y)/( log ^2x+ log^2y)=2 log y+(8 log x- log y)/( log^2x+ log^2y)=0

Kaylynn Cook

Kaylynn Cook

Answered question

2022-11-06

Logarithmic system of equations
Solve these equations
log x + log x + 8 log y log 2 x + log 2 y = 2
log y + 8 log x log y log 2 x + log 2 y = 0

Answer & Explanation

Hailee West

Hailee West

Beginner2022-11-07Added 15 answers

Let log x = u and log y = v. Your system becomes
u + u + 8 v u 2 + v 2 = 2 v + 8 u v u 2 + v 2 = 0
Put the 2 on the left side, and multiply by u 2 + v 2 :
u 3 + u v 2 2 u 2 2 v 2 + u + 8 v = 0 u 2 v + v 3 + 8 u v = 0
The resultant of the two left sides with respect to v is
260 u ( u 3 ) ( u + 1 ) ( u 2 2 u + 5 )
which must be 0 for a solution. With u = 0 we get v = 0 (not a solution to the original system). With u = 3 we get v = 2. With u = 1 we get v = 2. . With u = 1 ± 2 i (the roots of u 2 2 u + 5) we get v = ± 2 i.

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