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Thread: [SOLVED] Basic Logarithms

  1. #1
    Newbie adhiluhur's Avatar
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    Question [SOLVED] Basic Logarithms

    If $\displaystyle log_4(5)=-(3 over (2x))$ , find the value of $\displaystyle log_0.04(8)=$
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  2. #2
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    Hello, adhiluhur!

    I'm glad you solved it . . . Your problem is hard to read.


    $\displaystyle \text{If }\log_4(5) \:=\:-\frac{3}{2x}$ .[1]

    . . $\displaystyle \text{find the value of: }\:\log_{0.04}(8)$

    We have: .$\displaystyle \log_{0.04}(8) \:=\:P$ .[2]

    . . $\displaystyle (0.04)^P \:=\:8 \quad\Rightarrow\quad \left(\frac{1}{25}\right)^P \:=\:8 \quad\Rightarrow\quad \left(\frac{1}{5^2}\right)^P \:=\:8$

    . . $\displaystyle \left(5^{-2}\right)^P \:=\:8 \quad\Rightarrow\quad 5^{-2P} \:=\:8 $


    Take logs, base 4: .$\displaystyle \log_4\left(5^{-2P}\right) \:=\:\log_4(8)$

    . . $\displaystyle -2P\!\cdot\!\log_4(5) \:=\:\log_4\left(4^{\frac{3}{2}}\right) \quad\Rightarrow\quad -2P\!\cdot\!\log_4(5) \:=\:\frac{3}{2} $


    Substitute [1]: . $\displaystyle -2P\left(-\frac{3}{2x}\right) \:=\:\frac{3}{2} \quad\Rightarrow\quad P \:=\:\frac{x}{2}$ .[3]


    Equate [2] and [3]: . $\displaystyle \log_{.004}(8) \;=\;\frac{x}{2}$

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