SOLUTION: Solve for x: log(base 5)3x + log(base 5)(x-3) = 1 I see they both have a common base of 5. I'm not sure if I'm on the right track, but this is what I got so far: log(base 5)[3

Algebra ->  Algebra  -> Exponential-and-logarithmic-functions -> SOLUTION: Solve for x: log(base 5)3x + log(base 5)(x-3) = 1 I see they both have a common base of 5. I'm not sure if I'm on the right track, but this is what I got so far: log(base 5)[3      Log On

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Question 80522: Solve for x: log(base 5)3x + log(base 5)(x-3) = 1
I see they both have a common base of 5. I'm not sure if I'm on the right track, but this is what I got so far:
log(base 5)[3x(x-3)] = 1
3x(x-3) = 5^1
3x^2 - 9x =5
3x^2 - 9x - 5 = 0
I then tried factoring it out, but I don't think it works.
I think that first 3 in the problem might be throwing me off for some reason. Any help with this is greatly appreciated. Thanks.
Answer by Earlsdon(6103) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x:
Log%5B5%5D%283x%29%2BLog%5B5%5D%28x-3%29+=+1 Apply the product rule for logarithms.
Log%5B5%5D%28%283x%29%28x-3%29%29+=+1 Simplify the left side.
Log%5B5%5D%283x%5E2-9x%29+=+1 Rewrite in exponential form.
5%5E1+=+3x%5E2-9x Subtract 5 from both sides.
3x%5E2-9x-5+=+0 Your work up to this point is commendable!...and you are correct, this quadratic equation is not factorable, so you can use the quadratic formula: x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
x+=+%28-%28-9%29%2B-sqrt%28%28-9%29%5E2-4%283%29%28-5%29%29%29%2F2%283%29
x+=+%289%2B-sqrt%2881%2B60%29%29%2F6
x+=+%289%2B-sqrt%28141%29%29%2F6
The exact answers are:
x+=+3%2F2%2B%28sqrt%28141%29%29%2F6
x+=+3%2F2-%28sqrt%28141%29%29%2F6
The approximate answers are:
x+=+3.48 To the nearest hundredth.
x+=+-0.48 To the nearest hundredth.