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Mathematics 16 Online
OpenStudy (anonymous):

log base 5(3x+2)+log base 5(x-1)=1

OpenStudy (anonymous):

apply first laws of log at left side of the equation.... that is....\[\log_{5}m + \log_{5}n = \log_{5}mn \]

OpenStudy (anonymous):

then raise both side by 5... that will simplify the expression and you can solve for x....

OpenStudy (anonymous):

can you make it?

OpenStudy (anonymous):

\[5^{\log_{5}x }=x\]

OpenStudy (anonymous):

... 5 raise to log base 5 of x is simply equal to x... is just a sample you can reference to given equation above....

OpenStudy (anonymous):

sooo i got log base 5 (3x^2-x-2)=1 ....what do i do next??

OpenStudy (anonymous):

no you applied it wrong.... 5 ^ log base 5 will cancel to 1...

OpenStudy (anonymous):

to the right side... it would be 5^1 = 5

OpenStudy (anonymous):

so the equation will become... \[(3x+2)(x-1)=5\]

OpenStudy (anonymous):

you can now simply solve for x....

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