# Thread: Coverting to a single logarithm

1. ## Coverting to a single logarithm

i begin with this question
Express, log(x+1) - log(x-6) + 2 log x
as a single logarithm
so far i have come to an answer which is slightly puzzeling but i somehow mixed the expression to become equal to something which could be why i had the wrong answer, i have attempted this again, but im at a stage where im getting stuck, any chance for someone to correct me =]

workings
log(x+1/x-6) +logx^2
so now i have to times the two logs together
logx^2(x+1/x-6)
logx^2(x+1) divided by logX^2(x-6)......

workings
log(x+1/x-6) +logx^2
so now i have to times the two logs together
logx^2(x+1/x-6)
You are right up to this point

$\displaystyle \log (x+1) - \log (x-6) + 2 \log x = \log \left(\frac{x+1}{x-6}\right) + \log x^2 = \log \left(\frac{x^2(x+1)}{x-6}\right)$

i begin with this question
Express, log(x+1) - log(x-6) + 2 log x
as a single logarithm
so far i have come to an answer which is slightly puzzeling but i somehow mixed the expression to become equal to something which could be why i had the wrong answer, i have attempted this again, but im at a stage where im getting stuck, any chance for someone to correct me =]

workings
log(x+1/x-6) +logx^2
so now i have to times the two logs together
logx^2(x+1/x-6)
logx^2(x+1) divided by logX^2(x-6)......
You're adding the log(x^2) so you should have it in the numerator

$\displaystyle \log \left(\frac{x^2(x+1)}{x-6}\right)$