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Mathematics 22 Online
OpenStudy (anonymous):

Solve for x. Log[6] (5x+2) - log [6] (x-1) = log [6] (x-2)

OpenStudy (anonymous):

\[\LARGE \log_{6}(5x+2)-\log_6(x-1)=\log_6(x-2)\] this ?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

\[\log(a)-\log(b)=\log(\frac{a}{b})\] so start with \[\log(\frac{5x+2}{x-1})=\log(x-2)\] and therefore \[\frac{5x+2}{x-1}=x-2\]

OpenStudy (anonymous):

hah.. go satellite, your turn :F

OpenStudy (anonymous):

How exactly do I solve for x ?

OpenStudy (anonymous):

start from the last fraction satellite left for you... just use cross multiply and do the necessary simplification\multiply and you should find x... still having doubts ? :O

OpenStudy (anonymous):

A little bit. I don't want you guys to give me the answer but can you elaborate more or do a similar problem?

OpenStudy (anonymous):

\[\frac{5x+2}{x-1}=x-2\] \[5x+2=(x-2)(x-1)\] \[5x+2=x^2-3x+1\] now you have a quadratic equation to solve

OpenStudy (anonymous):

the answer is x=8

OpenStudy (anonymous):

ay... okay . thanks i got it from here.

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