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  1. #1
    Junior Member mathbuoy's Avatar
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    Talking LOgarithm proBLeM

    If log[a^2]b^2 - log[b]a = 4, find log[a]b.

    #characters in [ ] is at lower 'case' , attached to 'log'.


    *show working and explanation, pls...
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  2. #2
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    Use the formula
    $\displaystyle log_a(x)\;=\;\frac{ln(x)}{ln(a)}$
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  3. #3
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    Hello, mathbuoy!

    This is a strange one . . . I don't recommend natural logs.


    If $\displaystyle \log_{a^2}(b^2) - \log_b(a) \:=\: 4$, find $\displaystyle \log_a(b)$
    This is what I do when I encounter a strange base . . .

    Let $\displaystyle \log_{a^2}(b^2) \:=\:p \quad\Rightarrow\quad(a^2)^p = b^2 \quad\Rightarrow\quad a^{2p} = b^2$

    Take logs (base $\displaystyle a$): .$\displaystyle \log_a(a^{2p}) \:=\:\log_a(b^2) \quad\Rightarrow\quad 2p\log_a(a) \:=\:2\log_a(b) \quad\Rightarrow\quad p \:=\:\log_a(b)$

    . . Hence: .$\displaystyle \log_{a^2}(b^2) \;=\;\log_a(b) $


    And equation becomes: .$\displaystyle \log_a(b) - \log_b(a) \;=\;4$


    Use this identity on the second term: .$\displaystyle \log_b(a) \:=\:\frac{1}{\log_a(b)}$
    . . and we have: .$\displaystyle \log_a(b) - \frac{1}{\log_a(b)} \;=\;4$


    Multiply by $\displaystyle \log_a(b)\!:\quad \left[\log_a(b)\right]^2 - 1 \;=\;4\log_a(b) \quad\Rightarrow\quad\left[\log_a(b)\right]^2 - 4\log_a(b) - 1 \;=\;0$

    Quadratic Formula: .$\displaystyle \log_a(b) \;=\;\frac{4 \pm\sqrt{20}}{2}$

    . . Therefore: .$\displaystyle \log_a(b)\;=\;2 \pm\sqrt{5}$

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, mathbuoy!

    This is a strange one . . . I don't recommend natural logs.


    It works well anyway !
    We have to find $\displaystyle X\;=\;log_a(b)\;=\;\frac{ln(b)}{ln(a)}$

    $\displaystyle log_{a^2}(b^2)\;=\;\frac{ln(b^2)}{ln(a^2)}$

    $\displaystyle log_{a^2}(b^2)\;=\;\frac{2ln(b)}{2ln(a)}$

    $\displaystyle log_{a^2}(b^2)\;=\;\frac{ln(b)}{ln(a)}$

    $\displaystyle log_{a^2}(b^2)\;=\;X$

    Equation becomes
    X - 1/X = 4
    X - 4X - 1 = 0
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