Question

How does log conversion work? Why is it that I can express any log of base n as ln x / ln n?

Answer 1

Oh wait, I mean log base conversion, my bad.

Answer 2

I am not very good with this but this should help!

https://www.khanacademy.org/math/algebra2/exponential-and-logarithmic-functions/change-of-base-formula-for-logarithms/v/change-of-base-formula

Answer 3

Hope this helped! Have a great day!

Also a medal would be much appreciated! Just click best response next to my answer. Thank You!

@LanHikari22

Answer 4

we want to prove \(log_a b = \dfrac{log_c b}{log_c a}\)


to prove that:

\(y = log_a b\)
\(a^y = b\)
\(log_c (a^y) = log_c (b)\)
\(y \; log_c (a) = log_c (b)\)
\(y = \dfrac{log_c (b)}{ log_c (a) } \qquad = log_a b\) !!!!

Answer 5

@IrishBoy123 @Tom_Boy_Rebel Yeah! Thank you guys! I see this in principle but this still doesn't seem clear to me in a deductive reasoning point of view.
Let's say that we have log y = x. I can interpret this as how many bits, or digits, I need to represent y in base 10. ( Of course, this allows for root/fractional digits but yeah.) So if I want to convert this I can say ln y / ln 10 = x. But now isn't that interpreted as the amount of bits I need to represent y in e states divided by the amount of bits I need to interpret 10 in e? How is it that this indexing work, is what I'm trying to figure out.

Answer 6

no. it's just a logarithm. the rest is up to you :p

Answer 7

@IrishBoy123 Huh? What's that supposed to mean? Yeah. I know that it's a relationship, and I know that we proved the relationship. I'm trying to interpret the relationship now, though!

Answer 8

in higher math courses, there is only one log function ... not multiple log functions

Answer 9

@amistre64 Do you mean that they would cease to use other bases and stick with logs like e because it's compatible with change like in calculus, or that there's a completely different form for a function representing the whole?

Answer 10

all other logs can be defined as a scalar version of the natural log, as such there is the log function; and log(x) is defined as the natural log.

Answer 11

I think you are witting approximation for number of bits
i think you need the ceiling function after doing the log part

Answer 12

\[\lceil \log_n(x) \rceil=\lceil \frac{\ln(x)}{\ln(n)} \rceil \text{ is number of bits needed to } \\ \text{ represent } x \text{ in base } n\]
I believe

Answer 13

actually I think the ceiling function fails some times http://www.exploringbinary.com/number-of-bits-in-a-decimal-integer/
floor function of that log part then plus 1

Answer 14

i also I hope I'm using the word bits right
that is number of digits, right?

Answer 15

anyways I can't do better than Irish's proof of change of base formula

Answer 16

@freckles Yeah. I interpreted that, too. I guess I just need less... abstraction. I know the functionality, I have an interpretation, I need the inner workings of it. So I guess I'll dive into some number theory and see the relationships or something. Either ways, you're right, and I don't believe it actually fails. It's more of forming numbers differently.
For numbers like 1000, log 1000 is 3 because it can represent that in exactly 4 bits (3 and the 0th bit). But for numbers like 7, in binary for example, (111) log2 of 7 is 2.8 or somewhere around that, even though you really only need 2 bits. it doesn't have any concept of 111, it only has a leading one and then it, I'd guess, creates fractional bits that represent the fractional parts between b^n and b^n+1, everything is considered fractional and is represented with root bits in that region.