Solve for x: log(1/3)+log(x)=log(6x-5)

I got this to --log(1/3x)=log(6x-5)
but I'm not sure where to go from there.

Question

Solve for x: log(1/3)+log(x)=log(6x-5)

I got this to -->log(1/3x)=log(6x-5)
but I'm not sure where to go from there.

mathstudent55 · Answer

You did the first step correctly.
Now use this:

If $\log x = \log y$, then $x = y$.

stephanieelizzz · Answer

Not sure I get it. Would you be removing the logs since their bases are the same? @mathstudent55

mathstudent55 · Answer

Correct.
Using the same base log, if log A = log B then A = B.
You have log(1/3 x) = log(6x - 5), then it must follow that 1/3 x = 6x - 5

stephanieelizzz · Answer

Gotcha! For some reason I'm ending up with a completely different answer(5/6.3) from what I'm supposed to be getting (15/17). Maybe I'm solving it incorrectly? @mathstudent55

mathstudent55 · Answer

We'll do it together.

mathstudent55 · Answer

$\dfrac{1}{3}x = 6x - 5$

Multiply both sides by 3 to get rid of the denominator.

mathstudent55 · Answer

$x = 18x - 15$

Subtract 18x from both sides.

$-17x = -15$

Divide both sides by -17

$x = \dfrac{15}{17} $

stephanieelizzz · Answer

You're an angel. Thank you!!!

mathstudent55 · Answer

You're welcome.