SOLUTION: Can anyone help solve this? 1 - logx = log(3x-1) I tried: 1 - logx = log 3x - log 1 1 = logx + log 3x - log 1 1 = log4x - log 1 I also tried: 1 - logx = log(3x-1) div

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Can anyone help solve this? 1 - logx = log(3x-1) I tried: 1 - logx = log 3x - log 1 1 = logx + log 3x - log 1 1 = log4x - log 1 I also tried: 1 - logx = log(3x-1) div      Log On


   

Question 20730: Can anyone help solve this?
1 - logx = log(3x-1)
I tried:
1 - logx = log 3x - log 1
1 = logx + log 3x - log 1
1 = log4x - log 1
I also tried:
1 - logx = log(3x-1)
divide both sides by log
1-x = (3x-1)
2 - x = 3x
2 = 4x
1/2 = x
Or:
1 - logx = log(3x-1)
1 - 10 = log3x - log
-9 = 3logx - log
-9 = 3(10) - log
-9 = 30 - log
-39 = -log

The teacher's answer is 2
Can anyone help?
Thanks,
Sandy
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
1-logx+=+log%28%283x-1%29%29 Add logx to both sides of the equation.
1+=+log%28%283x-1%29%29+%2B+logx Apply the product rule for logarithms.
1+=+log%28%28%283x-1%29%2A%28x%29%29%29
1+=+log%28%283x%5E2-x%29%29 Recall that: If log%28a%29+=+1, then a+=+10 because: log%2810%29=1, therefore:
%283x%5E2-x%29+=+10 Subtract 10 from both sides.
3x%5E2-x-10=+0 Solve this quadratic equation by factoring.
%283x%2B5%29%28x-2%29+=+0 Apply the zero products principle.
%283x%2B5%29+=+0 and/or %28x-2%29+=+0
If 3x%2B5+=+0 then 3x+=+-5 and x+=+-5%2F3
If x-2+=+0 then x+=+2
The roots are:
x+=+2
x+=+%28-5%2F3%29
Check:
1) x = 2
1-log%282%29+=+log%28%283%282%29-1%29%29
1-0.30103+=+log%285%29
0.69897+=+0.69897
2) x+=+-5%2F3
1-log((-5/3)) = 0.77815..., -1.36437...
log(3(-5/3)-1) = 0.77815..., 1.36437...
x+=+%28-5%2F3%29 is not a valid root.
Answer is x = 2