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.Com
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) (Show Source): You can put this solution on YOUR website!
Add logx to both sides of the equation.
Apply the product rule for logarithms.
Recall that: If , then because: , therefore:
Subtract 10 from both sides.
Solve this quadratic equation by factoring.
Apply the zero products principle.
and/or
If then and
If then
The roots are:
Check:
1) x = 2
2)
1-log((-5/3)) = 0.77815..., -1.36437...
log(3(-5/3)-1) = 0.77815..., 1.36437...
is not a valid root.
Answer is x = 2
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