Substituting back into the original equation question:

Question

melissa_something · Answer

This is the question, I got x= 2, 12

melissa_something · Answer

My question is how could I make checking them easier on the test :/

anonymous · Answer

im actually not sure sorry!!

melissa_something · Answer

It's okay @Ethanwestbrooks

zepdrix · Answer

Make checking them easier?
To determine if any solutions are extraneous?$$\large\rm \log_2(-number)=bad$$Easy as that.
You don't need to do any further calculations.
x=2 is extraneous.

melissa_something · Answer

I thought it was only negative numbers and 0 that can't be true? How do you know?

zepdrix · Answer

Correct, the range of your log function is only positives,
no zero, no negative.$$\large\rm \color{red}{\log_2(2-6)}+\color{red}{\log_2(2-4)}-\log_2(2)$$We're getting log of negative value in both of these red spots, ya?

zepdrix · Answer

Sorry sorry, I mean the domain of your log function,
the values that can be `put inside of it`.

zepdrix · Answer

I colored them in red, do you not see the color?
The first one is log(-4) while the other is log(-2).
-4 and -2 are outside of the values we're allowed to put into the log function.

melissa_something · Answer

If you sent a picture or drawing I didn't get anything :(

zepdrix · Answer

When I do this:
$\large\rm \color{red}{\log_2(2-6)}+\color{red}{\log_2(2-4)}-\log_2(2)$

is the fancy math not showing up? :)

You must be using a cellphone or some nonsense XD lol

melissa_something · Answer

I refreshed and it worked! Lol

melissa_something · Answer

I see where you got the log (-4) but wouldnt the other one be log (-6) ?

zepdrix · Answer

2-6 = -6+2 = -4
this value becomes slightly `less negative` when we add positive 2.

2-4 = -4+2 = -2
again, becomes slightly less negative.

melissa_something · Answer

Oh my gosh, thank you. I get it. So did I just get lucky with this problem or do other problems require you to do it again?

zepdrix · Answer

Do what again? :o
Plug the other values in?

melissa_something · Answer

Yes, like for x=12, would I have to go back and do the same?

zepdrix · Answer

Yes you should plug in each solution and see if they hold true to the original equation:$$\large\rm \log_2(12-6)+\log_2(12-4)-\log_2(12)$$This gives us:$$\large\rm \log_2(+)~+~\log_2(+)-\log_2(+)$$

We don't get any negative funny business from x=12,
so we can finally conclude,$$\large\rm x=12$$$$\large\rm x\ne2$$

melissa_something · Answer

I got it, thank you so much @zepdrix <3

zepdrix · Answer

yay c:

melissa_something · Answer

You are a life saver! :D