How would I show that 1/log(subscript a) b = log(subscript b) a ?

I'm not sure where to start because of the different bases.

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- November 29th 2010, 02:02 PMTN17Is this logarithm true?
How would I show that 1/log(subscript a) b = log(subscript b) a ?

I'm not sure where to start because of the different bases. - November 29th 2010, 02:08 PMPlato
- November 29th 2010, 02:09 PMTN17
Yes, sorry for the typo.

- November 29th 2010, 02:20 PMTN17
- November 29th 2010, 02:20 PMPlato
You should know that .

So what does - November 29th 2010, 02:35 PMTN17
Hm... we haven't learned ln yet.

The teacher mentioned it really briefly in class, but didn't do a lesson on it. - November 29th 2010, 02:37 PMPlato
- November 29th 2010, 02:55 PMTN17
Couldbn't I say that

log (subscript b) a = log a / log b is true for any base/subscript.

Then 1 / [log (subscript a) b] = 1 / [log b / log a]

Where 1 / [log b / log a]

=log a / log b

Now use the fact at the top to change this into

log a / log b = log (subscript b) a

Therefore, 1 / [log (subscript a) b] = log (subscript b) a.

? - November 29th 2010, 02:57 PMPlato
You have the idea.

- November 29th 2010, 03:11 PMArchie Meade
- November 29th 2010, 03:24 PMTN17
I understand up to where you wrote a = b(superscript y)

I don't understand how you got a(superscript x)(superscript y) and onward... - November 29th 2010, 03:39 PMPlato
- November 29th 2010, 03:43 PMArchie Meade
- November 29th 2010, 05:08 PMSoroban
Hello, TN17!

Quote:

Let the right side equal .[1]

Then we have: .

Take logs, base

. . . . . .

Substitute [1]: .

- November 29th 2010, 05:33 PMTN17
Thanks. : )