Question

Log(3x+4) = 1 + log (x-6). Solve for x.

Answer 1

I moved both logs to one side so:
Log(3x+4) - log (x-6) = 1

and then I did that thing
\[\frac{ \log(3x+4) }{ \log(x-6) }=1\]

Answer 2

But I don't know what to do now...

Answer 3

That's good that you don't know what to do next, because what you did, did not work.

"That thing" should be this: \(\log(3x+4) - \log(x-6) = \log\left(\dfrac{3x+4}{x-6}\right)\).

Now, what's next?

Answer 4

That's what I have written down on paper I just wrote it wrong here, I swear. Thanks for pointing it out, though.

Answer 5

I know what I have to do, but I don't know HOW to do it. Like, if this was raised to an exponent I'd get the square root. I don't know what to do for logs...

Answer 6

My terminology is off, I know. I'm sorry.

Answer 7

Third time I've attempted this and I keep getting it wrong.

Answer 8

Mhmm?

Answer 9

log [(3x+4)/(x-6)] = 1

Answer 10

ask yourself when does the log of something =1 then solve the equation for that number

Answer 11

But I'm supposed to solve for x.

Answer 12

or to put it another way, what is the anti-log of 1?

Answer 13

\[10 = \frac{ 3x+4 }{ x-6 }\]

Answer 14

Very good Atlas!

Answer 15

okay let w= all the crap in the parenthesis log(w)=1 where does log of w=1 once you figure that out let that number=(3x+4)/(x-6)

Answer 16

and solve for x

Answer 17

I knew it was something simple. I hate myself. lol

Answer 18

Just forgot how logs worked.

Answer 19

Thanks everyone.

Answer 20

okay your welcome.

Answer 21

u r welcome :-)