# Relationship between the logs of a number with different bases

• Jan 7th 2011, 02:17 PM
BIOS
Relationship between the logs of a number with different bases
Can anyone explain step by step the relation between the logs of a number to different bases?

If we take a number x and do the following:

$\displaystyle log_{a}x$

then

$\displaystyle log_{b}x$

What is the relationship of the two?

I believe the formula that sums the relationship is:

$\displaystyle log_{b}x log_{a}x=log_{b}x$

Can anyone explain how this is reached?

P.S sorry for the formatting!
• Jan 7th 2011, 02:20 PM
e^(i*pi)
There is no relationship between them because the bases are different

If you said that $\displaystyle \log_b(x) = \log_a(x)$ then $\displaystyle a=b$ but as two expressions there is no relationship

You can use the change of base rule should you need to: $\displaystyle \log_a(x) = \dfrac{\log_b(x)}{\log_b(a)}$
• Jan 7th 2011, 02:25 PM
BIOS
Really? I have a book that states the following:

"The log of 0.278 is not equal to 1n 0.278. i.e. logarithms with different bases have different values. The different values are, however, related to eachother...."

:?
• Jan 7th 2011, 02:30 PM
Plato
Quote:

Originally Posted by BIOS
Really? I have a book that states the following:
"The log of 0.278 is not equal to 1n 0.278. i.e. logarithms with different bases have different values. The different values are, however, related to eachother...."

Of course that is the case.
If $\displaystyle a\ne b~\&~N\ne 1$ then $\displaystyle \log_a(N)\ne \log_b(N)$.
• Jan 7th 2011, 02:39 PM
BIOS
I think it's referring to a different relationship then i posted (and to example the book uses.)

E.g. the relationship between:

$\displaystyle \log_{b}a$ and the $\displaystyle \log_{a}x$

defining the relationship as:

$\displaystyle \log_{b}a\log_{a}x=\log_{b}x$

"This is the change of base formula which relates logarithms of a number relative to two different bases."

How is that the case if the logarithms are of two different numbers, namely a and x? or am i misunderstanding?
• Jan 7th 2011, 02:53 PM
Plato
Uning the definitions $\displaystyle \log_b(a)=\dfrac{\ln(a)}{\ln(b)}~\&~ \log_a(x)=\dfrac{\ln(x)}{\ln(a)}$ it is clear the by multiplying we

get $\displaystyle \,~~\log_b(a)\log_a(x)= \dfrac{\ln(x)}{\ln(b)}=\log_b(x)$,

• Jan 7th 2011, 03:03 PM
BIOS
I think i have it now. Thanks Plato. I was looking at the definition a different way.
• Jan 7th 2011, 03:38 PM
saravananbs
http://www.mathhelpforum.com/math-he...47aeaf1cd9.png=A(say)

then in exponetial form of this is x=a^A

http://www.mathhelpforum.com/math-he...28d00476c4.png=B(say)

then in exponetial form of this is x=b^B

then relation is a^A=b^B

i think 2 question is not valid always.
• Jan 7th 2011, 03:50 PM
Plato
Quote:

Originally Posted by saravananbs
http://www.mathhelpforum.com/math-he...47aeaf1cd9.png=A(say)
then in exponetial form of this is x=a^A
http://www.mathhelpforum.com/math-he...28d00476c4.png=B(say)
then in exponetial form of this is x=b^B then relation is a^A=b^B
i think 2 question is not valid always.

What a confusing posting.
Can you tell us what point you are trying to make?

We do know that $\displaystyle \log_b(a)=x$ is defined as $\displaystyle b^x=a$.