log 5x = log 3 + log (x-2) please help!!!

[ochocki]

I have always had a tough time with logs, here is the problem:

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


I notice that there is no base, this implies a base 10, so if all the bases are the same I assume i should go from there, I just don't know where to go Can someone give me a hint please?

I may have figured something out, tell me if I am on the right track:

for the right side of the equation:

log 3 + log (x-2)
log (3x-6)

which would leave me with:


log 5x = log (3x-6)

am i on the right track at all?

 

[ochocki]

Can anyone please help me?

 

[soroban]

Hello, ochocki!

You're doing great . . .

\(\displaystyle \log(5x) \:= \:\log(3)\,+\,\log(x\,-\,2)\)

I may have figured something out . . .

For the right side of the equation: \(\displaystyle \:\log(3)\,+\,\log(x\,-\,2)\:=\:\log[3(x\,-\,2)] \:=\:\log(3x\,-\,6)\)

which would leave me with: \(\displaystyle \:\log(5x) \:= \:\log(3x\,-\,6)\)

Am i on the right track at all? . Yes!

Now you can "un-log" both sides: \(\displaystyle \:5x\:=\:3x\,-\,6\) . . . etc.

 

[ochocki]

Arghh, I had a feeling I was doing it right, thanks for clarifying, i appreciate it.