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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- Nov 29th 2010, 01:02 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. - Nov 29th 2010, 01:08 PMPlato
- Nov 29th 2010, 01:09 PMTN17
Yes, sorry for the typo.

- Nov 29th 2010, 01:20 PMTN17
- Nov 29th 2010, 01:20 PMPlato
You should know that $\displaystyle \log_a(b)=\dfrac{\ln(b)}{\ln(a)}$.

So what does $\displaystyle \dfrac{\ln(a)}{\ln(b)}=~?$ - Nov 29th 2010, 01:35 PMTN17
Hm... we haven't learned ln yet.

The teacher mentioned it really briefly in class, but didn't do a lesson on it. - Nov 29th 2010, 01:37 PMPlato
- Nov 29th 2010, 01:55 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.

? - Nov 29th 2010, 01:57 PMPlato
You have the idea.$\displaystyle ~~\log_b(a)=\dfrac{\ln(a)}{\ln(b)}$

- Nov 29th 2010, 02:11 PMArchie Meade
- Nov 29th 2010, 02:24 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... - Nov 29th 2010, 02:39 PMPlato
- Nov 29th 2010, 02:43 PMArchie Meade
$\displaystyle a=b^y$

Now substitute in the value we have for $\displaystyle b$, which is $\displaystyle b=a^x$

$\displaystyle a=b^y=\left(a^x\right)^y$ since $\displaystyle b=a^x$

Using the laws of indices... $\displaystyle \left(c^p\right)^q=c^{pq}$

similar to $\displaystyle \left(2^2\right)^3=\left(2^2\right)\left(2^2\right )\left(2^2\right)=2^{2+2+2}=2^6=2^{(2)(3)}$

Therefore $\displaystyle \left(a^x\right)^y=a^{xy}$

but this is equal to $\displaystyle a$

hence $\displaystyle xy=1$

Although this is fine from the perspective of indices, Plato's method is the better one to get used to. - Nov 29th 2010, 04:08 PMSoroban
Hello, TN17!

Quote:

$\displaystyle \text{Show that: }\:\dfrac{1}{\log_a(b)} \:=\: \log_b(a)$

Let the right side equal $\displaystyle \,P\!:\;\;\log_b(a) \:=\:P$ .[1]

Then we have: .$\displaystyle b^P \:=\:a$

Take logs, base $\displaystyle a\!:\;\;\log_a(b^P) \:=\:\underbrace{\log_a(a)}_{\text{This is 1}}$

. . . . . . $\displaystyle P\cdot\log_a(b) \:=\:1 \quad\Rightarrow\quad \log_a(b) \:=\:\dfrac{1}{P}$

Substitute [1]: .$\displaystyle \log_a(b) \;=\;\dfrac{1}{\log_b(a)} $

- Nov 29th 2010, 04:33 PMTN17
Thanks. : )