How do I do this logarithmic equation?

Question

jojokiw3 · Answer

$$\log_{6} (x-1) + \log_{6} (x-2) = 2 \log_{6} \sqrt6 $$

zepdrix · Answer

log ruuuules :) yaaaaa

zepdrix · Answer

$$\Large\rm \log(a)+\log(b)=\log(ab)$$$$\Large\rm b \log(a)=\log(a^b)$$Looks like we'll need to use these two rules

jojokiw3 · Answer

$$x^2 - 3x + 2 = 1$$

jojokiw3 · Answer

?

zepdrix · Answer

=6 I think.
Hmm thinking about the step you got stuck on..

zepdrix · Answer

$$\Large\rm \log_6(stuff)=1$$Therefore,$$\Large\rm 6^1=stuff$$Yah?

jojokiw3 · Answer

$$2 \log_6 \sqrt 6 = \log_6 6 = 6^1 = 6?$$

zepdrix · Answer

No.$$2 \log_6 \sqrt 6 = \log_6 6 = 1$$

jojokiw3 · Answer

o.o where do I get the six from then?

zepdrix · Answer

But the other side of your equation is:$$\Large\rm \log_6(x^2-3x+2)$$That whole thing equals 1.$$\Large\rm \log_6(x^2-3x+2)=1$$You don't have x^2-3x+2=1 at this point.
You still have your log

zepdrix · Answer

You have to convert to exponential form to get rid of the log.

jojokiw3 · Answer

ahhh. I see what I missed. $$6^1 = x^2 - 3x +2$$

zepdrix · Answer

Yeaaa boyyyy, there we go!

jojokiw3 · Answer

Wheewww. I really need to get all these rules memorized completely. Thanks!

jojokiw3 · Answer

x = 4.

zepdrix · Answer

Because the x=-1 is extraneous? Ok looks good! :)