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Math Help - Simple Log Proofs

  1. #1
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    Simple Log Proofs

    1) Prove that $\log_{b} a + \log_{c} b + \log_{a} c = \displaystyle{\frac{1}{\log_{a} b}} + \displaystyle{\frac{1}{\log_{b} c}} + \displaystyle{\frac{1}{\log_{c} a}}$

    2) Prove that if $\log_{r} p = q$ and $\log_{q} r = p$, then $\log_{q} p = pq$

    If you could point me in the right direction I should be able to take it from there.

    Thanks again.
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  2. #2
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    Re: Simple Log Proofs

    log b (a) =x
    b^x =a
    loga((b^x)=loga (a)
    x(log a(b)=1
    x=1/loga(b)
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  3. #3
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    Re: Simple Log Proofs

    Quote Originally Posted by bjhopper View Post
    log b (a) =x
    b^x =a
    loga((b^x)=loga (a)
    x(log a(b)=1
    x=1/loga(b)
    @bjhopper, Why in the world do you not some effort to learn the code in LaTeX?

    \$\log_b(a)\$ gives $\log_b(a)$.
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  4. #4
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    Re: Simple Log Proofs

    Quote Originally Posted by bjhopper View Post
    log b (a) =x
    b^x =a
    loga((b^x)=loga (a)
    x(log a(b)=1
    x=1/loga(b)
    Damn. I thought it would be this, but I was weary of adding in an 'x'. Thank you.
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  5. #5
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    Re: Simple Log Proofs

    Hello, Fratricide!

    \text{(2) Prove that if }\:\begin{Bmatrix}\log_{r} p &=& q & [1] \\ \log_{q} r &=& p & [2]\end{Bmatrix},\;\text{then: }\:\log_{q} p \:=\: p\!\cdot\!q

    \text{From [1]: }\;\log_r(p) \,=\,q \quad\Rightarrow\quad p \:=\:r^q

    \text{Take logs, base }q\!:\;\;\log_q(p) \:=\:\log_q(r^q) \quad\Rightarrow\quad \log_q(p) \:=\:q\!\cdot\!\underbrace{\log_q(r)}_{\text{This is }p}
    \text{Therefore: }\;\log_q(p) \;=\;p\!\cdot\!q
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