# Explain why these steps were taken...

• Jun 14th 2009, 07:01 AM
maybealways
Explain why these steps were taken...
There is something which I don't quite understand, and I'm hoping someone could explain it to me.

Basically, I'm trying to find a statement that expresses this:
$
\log_{ab}x
$

in terms of p and q

$
Let,
$

$
\log_{a}x=p
$

$
\log_{b}x=q
$

$
\log_{a}x= \frac{\log_{ab}x}{\log_{ab}a}=p
$

$
\log_{b}x= \frac{\log_{ab}x}{\log_{ab}b}=q
$

I don't understand where the base ab came from in the above two expressions. Is there a way to do it without the ab?
• Jun 14th 2009, 07:10 AM
Showcase_22
Quote:

I don't understand where the base ab came from in the above two expressions
This is from the rules of changing the base of logs.

CHANGING THE BASE OF A LOGARITHM

Quote:

Is there a way to do it without the ab?
It depends on how you define a and b. In the way they are currently represented, they can only be expressed using $\log_{ab}$.
• Jun 14th 2009, 07:22 AM
maybealways
But then, wouldn't it be:

$
\log_{a}x=\frac{\log_x}{\log_a}
$

not

$
\log_{a}x=\frac{\log_{ab}x}{\log_{ab}a}
$
• Jun 14th 2009, 07:50 AM
Soroban
Hello, maybealways!

Quote:

Let: . $\begin{array}{c}\log_ax \:=\:p \\ \log_bx \:=\:q \end{array}$

Express $\log_{ab}x$ in terms of $p$ and $q$.

$\log_{a}x\:=\: \frac{\log_{ab}x}{\log_{ab}a}\:=\:p$

$\log_{b}x\:=\: \frac{\log_{ab}x}{\log_{ab}b}\:=\:q$

I don't understand where the base $ab$ came from.
Is there a way to do it without the $ab$?

Yes, but it requires some Olympic-level gymnastics.

We have: . $\log_ax \:=\:p \quad\Rightarrow\quad x \:=\:a^p \quad\Rightarrow\quad \:x^{\frac{1}{p}} \:=\:a$ .[1]

. . . .and: . $\log_bx \:=\:q \quad\Rightarrow\quad x \:=\:b^q \quad\Rightarrow\quad x^{\frac{1}{q}}\:=\:b$ .[2]

Mutliply [1] and [2]: . $x^{\frac{1}{p}}\cdot x^{\frac{1}{q}} \:=\:a\cdot b \quad\Rightarrow\quad x^{\frac{1}{p}+\frac{1}{q}}\:=\:ab$

$\text{Take logs (base }ab)\!:\;\;\log_{ab}\left(x^{\frac{1}{p} + \frac{1}{q}}\right) \:=\;\underbrace{\log_{ab}(ab)}_{\text{This is 1}}$

. . Then: . $\left(\tfrac{1}{p} + \tfrac{1}{q}\right)\log_{ab}(x) \:=\:1 \quad\Rightarrow\quad \log_{ab}(x) \:=\:\frac{1}{\frac{1}{p} + \frac{1}{q}}$

Therefore: . $\log_{ab}(x) \;=\;\frac{pq}{p+q}$