Proofs of Logarithms: Proving Equality of a and b

[thomasrules]

Can't start:

(log_{a}b)(log_{b}a) =1

 

[arildno]

Are you to prove that one?
Note that:
a^{\log_{a}(b)*\log_{b}(a)}=(a^{\log_{a}(b)})^{\log_{b}(a)}
Can you continue?

 

[thomasrules]

how'd u get that

 

[thomasrules]

You know that a^{\log_{a}(b)} = a.

how'd u get a^{\log_{a}(b)} = a.

 

[courtrigrad]

It should be a^{\log_{a}(b)} = b. My fault.

 

[thomasrules]

i'm sorry but i don't even understand that step from the original to that

 

[courtrigrad]

You get a^{\log_{a}(b)*\log_{b}(a)} = a^{1} therefore (log_{a}b)(log_{b}a) =1. Basically, you start with the base a and raise it to the respective powers on the left and right hand side of the equation. You could have used b as the base instead.

Also a^{\log_{a}(b)} = b. Look at an example. 10^{\log_{10}(100)} = 10^{2} = 100 = b

 

[thomasrules]

so then:

log_{a}a=(log_{a}b)(log_{b}a)

?

 

[courtrigrad]

\log_{a}a = 1 is true. But that is not how I showed that (\log_{a}b)(\log_{b}a) =1. We have a^{\log_{a}(b)*\log_{b}(a)} = a^{1}. Since the bases of both sides of the equation is a, we can equate their exponents to each other. That means (\log_{a}b)(\log_{b}a) =1.

 

[thomasrules]

but that's not the proof...

 

[arildno]

how'd u get that

By the normal rule for a product in the exponent.

 

[lkh1986]

We can change the base for the second term, which gives
log (base b) a = log (base a ) a / log (base a) b

From there, we can cancel out the term log (base a) b and only log (base a ) a remains, which gives the answer of 1.

 

[The Bob]

Try looking at the 2nd post.

Note that:
a^{\log_{a}(b)*\log_{b}(a)}=(a^{\log_{a}(b)})^{\log_{b}(a)}

Here is the proof that you want. As arildno said, continue this and you will find that (log_{a}b)(log_{b}a) =1, which is what you said you wanted. All you need to use is the laws of indices, the idea of how logs work and you are done, no joke.

All the best,

The Bob (2004 ©)

 

[HallsofIvy]

Can't start:

(log_{a}b)(log_{b}a) =1

Get them to the same base. Remember that log_a b= x means that b= a^x. Now take the logarithm, to base b of both sides of that:
log_b b= 1= log_b a^x= x log_{b[/sup]a
Since x= loga b, ...}