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Math Help - Log problem again

  1. #1
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    Post Log problem again

    Hi all, I have this log problem that I having trouble getting \log_{8} \frac {x}{2} = \frac {\log_{8}x}{\log_{8}2} can someone please help me solve this log equation.
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  2. #2
    Member kjchauhan's Avatar
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    By changing the base of log..

    \frac{\log_{e}\frac{x}{2}}{\log_{e}{8}}=\frac{\log  _{e}x}{\log_{e}2}

    \therefore\frac{\log_{e}\frac{x}{2}}{3\log_{e}{2}}  =\frac{\log_{e}x}{\log_{e}2}


    \therefore\log_{e}x-\log_{e}2=3\log_{e}x

    \therefore\log_{e}x-\log_{e}2=\log_{e}x^3

    \therefore\log_{e}x-\log_{e}x^3=\log_{e}2

    \therefore\log_{e}\frac{x}{x^3}=\log_{e}2

    \therefore x=2x^3

    \therefore 2x^3-x=0

    Now solve this Equation....
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  3. #3
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    Hello, scrible!

    Solve for x\!:\;\;\log_8\left(\frac {x}{2}\right) \;=\; \frac{\log_8(x)}{\log_8(2)}
    First, evaluate \log_8(2)

    Let: . \log_8(2) \:=\:P \quad\Rightarrow\quad 8^P \:=\:2 \quad\Rightarrow\quad \left(2^3\right)^P \:=\:2 \quad\Rightarrow \quad 2^{3P} \:=\:2^1

    Hence: . 3P \:=\:1 \quad\Rightarrow\quad P \:=\:\tfrac{1}{3}

    Therefore: . \log_8(2) \,=\,\tfrac{1}{3} .[1]


    The equation is: . \log_8(x) - \log_8(2) \;=\;\frac{\log_8(x)}{\log_8(2)}

    Substitute [1]: . \log_8(x) - \tfrac{1}{3} \;=\;\frac{\log_8(x)}{\frac{1}{3}} \quad\Rightarrow\quad \log_8(x) - \tfrac{1}{3} \;=\;3\log_8(x)

    Multiply by 3: . 3\log_8(x) - 1 \;=\;9\log_8(x) \quad\Rightarrow\quad -6\log_8(x) \;=\;1 \quad\Rightarrow\quad \log_8(x) \;=\;-\tfrac{1}{6}

    Hence: . x \;=\;8^{-\frac{1}{6}} \;=\;\left(2^3\right)^{-\frac{1}{6}} \;=\;2^{-\frac{1}{2}}

    Therefore: . \boxed{x \;=\;\frac{1}{\sqrt{2}}}

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

    First, evaluate \log_8(2)

    Let: . \log_8(2) \:=\:P \quad\Rightarrow\quad 8^P \:=\:2 \quad\Rightarrow\quad \left(2^3\right)^P \:=\:2 \quad\Rightarrow \quad 2^{3P} \:=\:2^1

    Hence: . 3P \:=\:1 \quad\Rightarrow\quad P \:=\:\tfrac{1}{3}

    Therefore: . \log_8(2) \,=\,\tfrac{1}{3} .[1]


    The equation is: . \log_8(x) - \log_8(2) \;=\;\frac{\log_8(x)}{\log_8(2)}

    Substitute [1]: . \log_8(x) - \tfrac{1}{3} \;=\;\frac{\log_8(x)}{\frac{1}{3}} \quad\Rightarrow\quad \log_8(x) - \tfrac{1}{3} \;=\;3\log_8(x)

    Multiply by 3: . 3\log_8(x) - 1 \;=\;9\log_8(x) \quad\Rightarrow\quad -6\log_8(x) \;=\;1 \quad\Rightarrow\quad \log_8(x) \;=\;-\tfrac{1}{6}

    Hence: . x \;=\;8^{-\frac{1}{6}} \;=\;\left(2^3\right)^{-\frac{1}{6}} \;=\;2^{-\frac{1}{2}}

    Therefore: . \boxed{x \;=\;\frac{1}{\sqrt{2}}}
    Thank you, but when I checked the answer in the book they has the answer as being \frac{\sqrt{2}}{2}
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  5. #5
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    Quote Originally Posted by scrible View Post
    Thank you, but when I checked the answer in the book they has the answer as being \frac{\sqrt{2}}{2}
    Formally, you should never have a square root in the denominator. Here's what the book did (multiply by 1) to get rid of the square root in the bottom.

    x = \frac{1}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}<br />
= \frac{\sqrt{2}}{2}
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