# Thread: Explain why these steps were taken...

1. ## 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:
$\displaystyle \log_{ab}x$
in terms of p and q

$\displaystyle Let,$
$\displaystyle \log_{a}x=p$
$\displaystyle \log_{b}x=q$

$\displaystyle \log_{a}x= \frac{\log_{ab}x}{\log_{ab}a}=p$
$\displaystyle \log_{b}x= \frac{\log_{ab}x}{\log_{ab}b}=q$

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?

2. 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

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 $\displaystyle \log_{ab}$.

3. But then, wouldn't it be:

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

not

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

4. Hello, maybealways!

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

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

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

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

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

Yes, but it requires some Olympic-level gymnastics.

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

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

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

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

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

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