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Math Help - finding logs in terms of another log

  1. #1
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    Unhappy finding logs in terms of another log

    I did several of these but am confused by this one.

    let a= log2 5 and b= log3 2. Express the following in terms of a, b, or both.

    log2 6=? and log5 3=?

    cancel the first one, I figured it out. I am now trying (and trying) the log5 3= (and that 5 is supposed to be the base of the log...I just realized that there's html math commands but haven't used them yet.)
    Last edited by jlefholtz; March 18th 2009 at 11:36 AM.
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  2. #2
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    Quote Originally Posted by jlefholtz View Post
    I did several of these but am confused by this one.

    let a= log2 5 and b= log3 2. Express the following in terms of a, b, or both.

    log2 6=? and log5 3=?
    1. b = \log_3(2)~\implies~\dfrac1b = \log_2(3)

    \log_2(6)=\log_2(2\cdot 3)=\log_2(2)+\log_2(3)=1+\dfrac1b

    2. \log_5(3)=\log_5\left(\dfrac62\right)=\log_5(6)-\log_5(2) = \dfrac{\log_2(6)}{\log_2(5)}-\dfrac1a=\dfrac{1+\dfrac1b}a-\dfrac1a
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  3. #3
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    thanks!

    Oh thanks! I was going to try using division in the second one but didn't know if it was breaking a rule of logarithms. I can do it from here!
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  4. #4
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    Hello, jlefholtz!

    These require some Olympic-level gymnastics . . .


    Let: . a\:=\:\log_2(5)\,\text{ and }\,b\:=\:\log_3(2)

    Express the following in terms of a and/or b.

    (1)\;\log_2(6)

    (2)\;\log_5(3)
    Theorem: . \log_b(a) \:=\:\frac{1}{\log_a(b)}


    (1) We have: . \log_2(6) \:=\:\log_2(3\cdot2) \:=\:\log_2(3) + \log_2(2)

    . . . . Hence: . \log_2(6) \:=\:\log_2(3) + 1 .[1]


    . . .We are given: . \log_3(2) \:=\:b \quad\Rightarrow\quad \frac{1}{\log_2(3)} \:=\:b \quad\Rightarrow\quad \log_2(3) \:=\:\frac{1}{b}

    . . .Substitute into [1]: . \log_2(6) \:=\:\frac{1}{b} + 1 \quad\Rightarrow\quad\boxed{ \log_2(6)\;=\;\frac{1+b}{b}}



    (2) We are given: . \begin{array}{ccccccc}\log_2(5) \:=\:a & \Longrightarrow & 2^a \:=\: 5 & {\color{blue}[2]}\\ \log_3(2) \:=\: b & \Longrightarrow & 3^b \:=\: 2 & {\color{blue}[3]}\end{array}

    Raise [3] to the power a\!:\;\;(3^b)^a \:=\:2^a \quad\Rightarrow\quad 3^{ab} \:=\:2^a

    But from [1]: . 2^a \:=\:5

    . . So we have: . 3^{ab} \:=\:5


    \text{Take logs (base 3): }\;\log_3(3^{ab}) \:=\:\log_3(5) \quad\Rightarrow\quad ab\cdot\underbrace{\log_3(3)}_{\text{This is 1}} \:=\:\log_3(5)

    We have: . \log_3(5) \:=\:ab \quad\Rightarrow\quad \frac{1}{\log_5(3)} \:=\:ab \quad\Rightarrow\quad \boxed{\log_5(3) \:=\:\frac{1}{ab}}


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