SOLUTION: Hi, I need help in solving this logarith problem:- Solve the simultaneous equations log(x+y)= 2log(x) (this equation doesn't have any base) log(y)= log(2)+ log(x-1) My

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Hi, I need help in solving this logarith problem:- Solve the simultaneous equations log(x+y)= 2log(x) (this equation doesn't have any base) log(y)= log(2)+ log(x-1) My       Log On


   

Question 486725: Hi, I need help in solving this logarith problem:-
Solve the simultaneous equations
log(x+y)= 2log(x) (this equation doesn't have any base)
log(y)= log(2)+ log(x-1)
My attempt:
I simplified the first line :-
log(x+y)= 2log(x)
=log(x,x+y)= log(x,x^2) (I put the (x) as a base)
=log(x,x+y)= 2 (I simplified log(x,x^2) to number 2)
= x^2= x+y
Similarly I simplified the second line to be (y = 2x-2)
I don't know what to do any further now.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Your work so far is good!
1) x%5E2+=+x%2By and...
2) y+=+2x-2 Substitute this for y in equaton 1) to get...
1a) x%5E2+=+x%2B%282x-2%29 Simplify and solve for x:
1b) x%5E2-3x%2B2+=+0 Solve by factoring:
1c) %28x-1%29%28x-2%29+=+0 so...
1d) highlight%28x+=+1%29 or highlight%28x+=+2%29 Now plug each of these, in turn, in equation 2 and solve for y.
2a) y+=+2x-2 Substitute x+=+1:
2b) highlight%28y+=+0%29 and...
2c) y+=+2x-2 Substitute x+=+2:
2d) highlight%28y+=+2%29
So there are four possible solutions:
(1, 0), (1, 2), (2, 0), (2, 2)
Let's check them all:
log%28x%2By%29+=+2Log%28x%29 Substitute x+=+1 and y+=+0
Log%281%29+=+2Log%281%29
0+=+0 This solution is valid!
Try the other solutions yourself and see what happens!