No I didn't mean to write x as the baseIf the x represents the base, then logx(abc) = logx(a) + logx(b) + logx(c)
Well that is one way to have the question make sense.No I didn't mean to write x as the base
Fine, can you please tell us what you did mean?No I didn't mean to write x as the base
I meant log base a of x equals 2 and so onFine, can you please tell us what you did mean?
Maybe you meant (log(a))x=2, (log(b))x=3 and (log(c)x)=6. You need to tell us what you are taking the logs of.
I don't know mabye I should try to solve the other way around because this really doesn't make senseSo we have \(\log_{a}(x)=2,~\log_{b}(x)=3,~\&~\log_{c}(x)=6\)
OR \(x=a^2,~x=b^3~\&~x=c^6\)
Now what?
Actually the question as written makes no sense.I don't know mabye I should try to solve the other way around because this really doesn't make sense
I understand this to meanIf log(a)x=2, log(b)x=3 and log(c)x=6 then how much is log(abc)x going to be?
Can you explain it a bit more cause I don't get it.I understand this to mean
If loga(x)=2, logb(x)=3, and logc(x)=6 then how much is logabc(x) going to be?
When I see an unknown as the base of a log, I immediately think about the fact that logx(y) = 1/logy(x). Are you familiar with that? (It can be easily proved using the change of base formula.)
Use that to rewrite all four equations in the problem, and you'll find that the answer is easy!
(Using pka's idea in post #8, you can express a, b, and c in terms of x, and then express abc likewise. The answer again drops out naturally.)
Can you explain it a bit more cause I don't get it.
No I don't know this formulaPlease make an attempt and show your work. It's a lot easier to guide someone when you can see where they are not understanding.
If nothing else, tell me whether you know the change of base formula.
I will tell you this answer if you tell us if you reviewed the original question to be clear that you posted it correctly.No I don't know this formula
Yeah I checked the question and it is posted correctlyI will tell you this answer if you tell us if you reviewed the original question to be clear that you posted it correctly.
If \(T=\log_b(S)\) then we can change the base to \(T=\dfrac{\log(S)}{\log(b)}\)
Example \(\log_{10}(75)=\dfrac{\log(75)}{\log(10)}\).
So what Prof Peterson means: if \(\log_x(y)=\dfrac{\log(y)}{\log(x)}=\dfrac{1}{\frac{\log(x)}{\log(y)}}=\dfrac{1}{\log_y(x)}\)
Yes, I knew it was posted correctly, because it has a very nice answer.
Now please show SOME bit of work so we can tell what you CAN do, rather than just what you can't. Try rewriting each equation in terms of a base-x log, and show us what you get. Also think about what you might be able to do with the results.