What if you're expanding a logarithm, and you get 1 - log(3,(x)) + 5, do you add 1 and 5?

Question

anonymous · Answer

If you're in the process of expanding a logarithm, and you get 1 - log(3,(x)) + 5
Do you combine like terms, such as the 1 and 5? Or do you leave it like that?

zepdrix · Answer

1-log(3,x) + 5? Hmm yah you can combine the 1 and 5 if they're both outside of the logarithm.

Do you have a specific problem in mind? I just want to make sure I'm not misunderstanding you <:O

anonymous · Answer

Sorry no specifics, but are you sure? Then how would you expand again? Because that 1 came from log(3) 3 and that 5 came from log(5)125

anonymous · Answer

I just simplified it

zepdrix · Answer

Oh ok I see what you did :) So 3^1 = 3, so that simplified to 1.
Hmm, 5^3=125, so i think the other one simplifies to 3, yes? :o
And then yes you can combine them in the way that you wanted :D shouldn't be any problem.

anonymous · Answer

Are you 200% sure?

anonymous · Answer

Because it doesn't make sense how i'll re-expand it

zepdrix · Answer

$$2=\log_{10}10 0$$
$$2=\log_{3} 9$$
There isn't a single way to re-expand it. As you can see, there are many logs that give us a value of 2.

I think maybe the idea with this problem was to combine the logs? write them all as one log? If that's the case, we can certainly do that :)

anonymous · Answer

It was to expand it as much as possible

zepdrix · Answer

$$\log_{3}3-\log_{3}x=\log_{3}(\frac{ 3 }{ x })$$$$\log_{5}125=3\rightarrow 3=\log_{3}27$$
$$\log_{3}3-\log_{3}x+\log_{5}125=\log_{3}(\frac{ 3*27 }{ x })=\log_{3}(\frac{ 81 }{ x })$$
Hmm I'm sorry maybe I'm misunderstanding what you're suppose to be doing with this problem :(
You can combine all the logs together like this, I'm not sure if that's what you're trying to do though.