SOLUTION: log[10](x)+log[10](3x+55)=2 How do you solve this?

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Question 722979: log[10](x)+log[10](3x+55)=2
How do you solve this?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
log%2810%2Cx%29%2Blog%2810%2C%283x%2B55%29%29=2
Since a sum of logs (on the same base) is the log of the product, we can re-write that as
log%2810%2C%28x%283x%2B55%29%29%29=2 --> log%2810%2C%283x%5E2%2B55x%29%29=2 --> 3x%5E2%2B55x=10%5E2 --> 3x%5E2%2B55x=100 --> 3x%5E2%2B55x-100=0

The quadratic equation 3x%5E2%2B55x-100=0 can be easily solved by factoring or by using the quadratic formula.

Factoring:
3x%5E2%2B55x-100=0 --> %283x-5%29%28x%2B20%29=0 --> x=-20 or x=5%2F3

Using the quadratic formula:
x=%28-55+%2B-+sqrt%2855%5E2-4%2A3%2A%28-100%29%29%29%2F%282%2A3%29 --> x=%28-55+%2B-+sqrt%283025%2B1200%29%29%2F6 --> x=%28-55+%2B-+sqrt%284225%29%29%2F6 --> x=%28-55+%2B-+65%29%2F6 --> x=-120%2F6=-20 or x=10%2F6=5%2F3

x=-20 is not a solution of log%2810%2Cx%29%2Blog%2810%2C%283x%2B55%29%29=2
(It is an extraneous solution that must be eliminated, because log%2810%2C%28-20%29%29 does not exist).

The solution is highlight%28x=5%2F3%29