Results 1 to 4 of 4

Thread: Explain why these steps were taken...

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    18

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    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 $\displaystyle \log_{ab}$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2009
    Posts
    18
    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}
    $
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    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: .$\displaystyle \log_{ab}(x) \;=\;\frac{pq}{p+q}$

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Find X? show and explain steps to find x
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Mar 30th 2011, 06:24 PM
  2. Replies: 2
    Last Post: Sep 26th 2010, 02:45 AM
  3. Replies: 1
    Last Post: Feb 8th 2010, 05:31 PM
  4. I want to explain the latest steps from this Q
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Dec 25th 2009, 01:24 AM
  5. Replies: 1
    Last Post: Sep 19th 2009, 01:12 AM

Search Tags


/mathhelpforum @mathhelpforum