Logarithms Problem

[Yavor Apostolov]

Hello, I seem to be stuck on the following problem:

If log12(27) = a, then using a, log12(18) = ?

Any help would be greatly appreciated.

 

[tkhunny]

[math]log_{12}(18) = log_{12}\left(2\cdot 3^{2}\right) = log_{12}(2) + 2\cdot log_{12}(3)[/math]
[math]a = log_{12}(27) = log_{12}\left(3^{3}\right) = 3\cdot log_{12}(3) = a[/math]
Does any of that get us anywhere?

Maybe you already looked at that and didn't see much progress. If so, you should have put it in the very first post.

 

[Yavor Apostolov]

Right. I have indeed tried that, but got only to log12(2) + 2/3a, but I can't figure out how to express log12(2).

 

[Deleted member 4993]

Right. I have indeed tried that, but got only to log12(2) + 2/3a, but I can't figure out how to express log12(2).

Please share your WORK (not only the last step). Looking at your work will help us guide you to correct solution.

 

[Steven G]

Hello, I seem to be stuck on the following problem:

If log12(27) = a, then using a, log12(18) = ?

Any help would be greatly appreciated.

Let b = log₁₂(18)

Now 12^b = 18 and 12^a = 27

12^b/12^a = 18/27 = 2/3

12^b = (2/3)12^a

Now solve for b

 

[Steven G]

Right. I have indeed tried that, but got only to log12(2) + 2/3a, but I can't figure out how to express log12(2).

What is wrong with log₁₂(18) = log₁₂(2) + (2/3)a as your answer? What form are you looking to write log₁₂2 in???