SOLUTION: Solve the simultaneous equations : logbase2(x) + logbase8(y) = -1 logbase4(x) + logbase2(y) = 2 Cannot solve, spent hours on it Any help would be appreciated

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve the simultaneous equations : logbase2(x) + logbase8(y) = -1 logbase4(x) + logbase2(y) = 2 Cannot solve, spent hours on it Any help would be appreciated      Log On


   

Question 565979: Solve the simultaneous equations :

logbase2(x) + logbase8(y) = -1
logbase4(x) + logbase2(y) = 2

Cannot solve, spent hours on it
Any help would be appreciated
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the simultaneous equations :
logbase2(x) + logbase8(y) = -1
logbase4(x) + logbase2(y) = 2
**
Change to base2:
log2(x) + log2(y)/log2(8) = -1
log2(x)/log2(4) + log2(y) = 2
..
log2(4)=2
log2(8)=3
..
log2(x) + log2(y)/3= -1
log2(x)/2 + log2(y) = 2
..
3log2(x) + log2(y)= -3
log2(x) +2 log2(y) =4
..
6log2(x) + 2log2(y)= -6
log2(x) +2 log2(y) =4
subtract
6log2(x)-log2(x)=-10
log2(x^6)-log2(x)=-10
log2(x^6/x)=-10
log2(x^5)=-10
convert to exponential form: base(2) raised to log of number(-10)=number(x^5)
2^-10=x^5
take 5th root of both sides
x=(2^-2)=1/(2^2)=1/4
..
solving for y
log2(x) +2 log2(y) =4
log2(2^-2)+2log2(y)=4
-2+2log2(y)=4
2log2(y)=6
log2(y)=3
y=8
ans:
x=1/4
y=8