Question

logx + log(x-2) = log3x
The first step is to get everything on one side so..
logx + log(x-2) - log3x = 0
What do I do next?

Answer 1

You have my attention ^_^
Now... all these logs have the same base, yes?

Answer 2

Yeah, base of x right?

Answer 3

Base of x? I don't think so.
In fact, I don't think it would matter what base these logs have, as long as it's the same.
Do you see a small number underneath the logs like this:
\[\Large \log_{\color{red}b}x\]
or not?

Answer 4

Oh, there's no number.

Answer 5

\[\large logx + \log(x-2) =\log3x\]
that's exactly how it is.

Answer 6

Well then, in that case, this can be solved rather quickly and easily, in two ways, in fact ^_^

First, take advantage of this property:
\[\Large \log(MN) = \log(M) + \log(N)\]

Answer 7

So log x(x-2) = log 3x

Answer 8

I guess you could do that... and then?

Answer 9

Since there is no base (the small number) it is generally assumed that the base is 10. (I think)
Regardless, raise 10 to both sides of the equation... like so:
\[\Large \log(M) = \log(N)\\\Large 10^{\log(M)}= 10^{\log(N)}\\\Large \ \\\Large M = N\]

Answer 10

log x(x-2) = log 3x
divide by log 3x?..
log x(x-2) / log 3x = 1
x(x-2)/3x = 1
(x-2)/3 = 1
x-2 = 3
x = 5

Answer 11

Answer is correct, but your third line is erroneous ^_^

It should instead be
x(x-2)/ 3x = 1 right away, or
log [x(x-2)/3x] = 0

But anyway...
Here's another way, which, I believe, is less messy :)
\[\large \log(x) + \log(x-2)=\log(3x)\]\[\large \log(x)+\log(x-2) = \color{blue}{\log(3) + \log(x)}\]

Cancel out log(x) from both sides, we get...
\[\Large \log(x-2) = \log(3)\]\[\Large x-2= 3\]\[\Large x = 5 \color{green}{\qquad\checkmark }\]

Answer 12

Thank you!
That makes more sense lol.
Forgot about how you can turn log (3x) into log (3) + log (x) .

Answer 13

That doesn't always work, mind you, the way you did it was more...general, save for that tiny error in the middle.

Nice work, anyway. ^_^