Is this logarithm true?

**TN17** · November 29th 2010, 01:02 PM

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.

**Plato** · November 29th 2010, 01:08 PM

Originally Posted by TN17

How would I show that 1/logab = logba ?
I'm not sure where to start because of the different bases.

Is this the question?
$\log_a(b)=\dfrac{1}{\log_b(a)}$

**TN17** · November 29th 2010, 01:09 PM

Yes, sorry for the typo.

**TN17** · November 29th 2010, 01:20 PM

Originally Posted by Plato

Is this the question?
$\log_a(b)=\dfrac{1}{\log_b(a)}$

Yes, that's what I meant. Sorry for the typo.

**Plato** · November 29th 2010, 01:20 PM

You should know that $\log_a(b)=\dfrac{\ln(b)}{\ln(a)}$ .
So what does $\dfrac{\ln(a)}{\ln(b)}=~?$

**TN17** · November 29th 2010, 01:35 PM

Hm... we haven't learned ln yet.
The teacher mentioned it really briefly in class, but didn't do a lesson on it.

**Plato** · November 29th 2010, 01:37 PM

Originally Posted by TN17

Hm... we haven't learned ln yet.
The teacher mentioned it really briefly in class, but didn't do a lesson on it.

Well that is the proof.

**TN17** · November 29th 2010, 01:55 PM

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.

?

**Plato** · November 29th 2010, 01:57 PM

You have the idea. $~~\log_b(a)=\dfrac{\ln(a)}{\ln(b)}$

**Archie Meade** · November 29th 2010, 02:11 PM

Originally Posted by TN17

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.

Alternatively,

$log_ab=x\Rightarrow\ b=a^x$

$log_ba=y\Rightarrow\ a=b^y=\left(a^x\right)^y=a^{xy}\Rightarrow\ xy=1$

$\Rightarrow\ y=\frac{1}{x}\Rightarrow\ log_ba=\frac{1}{log_ab}$

**TN17** · November 29th 2010, 02:24 PM

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...

**Plato** · November 29th 2010, 02:39 PM

Originally Posted by TN17

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...

Your basic problem is not knowing or understanding the basic definitions.

$\log_b(a)=t\text{ if and only if }a=b^t,~~a>0~\&~b>0.$

**Archie Meade** · November 29th 2010, 02:43 PM

Originally Posted by TN17

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...

$a=b^y$

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

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

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

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

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

but this is equal to $a$

hence $xy=1$

Although this is fine from the perspective of indices, Plato's method is the better one to get used to.

**Soroban** · November 29th 2010, 04:08 PM

Hello, TN17!

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

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

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

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

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

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

**TN17** · November 29th 2010, 04:33 PM

Thanks. : )

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