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Thread: Log problem

  1. #1
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    Log problem

    prove that $\displaystyle f(-x)=f(x) $ if $\displaystyle f(x)=(x-\log_3 \frac{2+x}{2-x}) \cdot \log_2 \frac{x+1}{x-1}$

    $\displaystyle f(-x)=(-x-\log_3 \frac{2+(-x)}{2-(-x)}) \cdot \log_2 \frac{-x+1}{-x-1}$

    $\displaystyle f(-x)=(-x-\log_3 \frac{2-x}{2+x}) \cdot \log_2 \frac{-(x-1)}{-(x+1)}$

    $\displaystyle f(-x)=(-x-\log_3 \frac{2-x}{2+x}) \cdot \log_2 \frac{x-1}{x+1}$

    ???
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by livmed View Post
    prove that $\displaystyle f(-x)=f(x) $ if $\displaystyle f(x)=(x-\log_3 \frac{2+x}{2-x}) \cdot \log_2 \frac{x+1}{x-1}$

    $\displaystyle f(-x)=(-x-\log_3 \frac{2+(-x)}{2-(-x)}) \cdot \log_2 \frac{-x+1}{-x-1}$

    $\displaystyle f(-x)=(-x-\log_3 \frac{2-x}{2+x}) \cdot \log_2 \frac{-(x-1)}{-(x+1)}$

    $\displaystyle f(-x)=(-x-\log_3 \frac{2-x}{2+x}) \cdot \log_2 \frac{x-1}{x+1}$

    ???
    This is an exercise in the properties of logs. Try to reduce the logs to an expression containing one logarithm only. Make use of the facts that $\displaystyle \log{a}-\log{b}=\log\frac{a}{b}$ and the other properties that may be needed.
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  3. #3
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    Hello, livmed!

    $\displaystyle \displaystyle\text{If }f(x)\:=\:\left(x-\log_3 \frac{2+x}{2-x}\right) \cdot \log_2 \frac{x+1}{x-1},\;\text{prove that }f(\text{-}x) \,=\,f(x).$


    $\displaystyle \displaystyle f(\text{-}x)\;=\;\left(\text{-}x-\log_3 \frac{2+(\text{-}x)}{2-(\text{-}x)}\right) \cdot \log_2 \frac{\text{-}x+1}{\text{-}x-1}$

    $\displaystyle \displaystyle f(\text{-}x)\;=\;\left(-x-\log_3 \frac{2-x}{2+x}\right) \cdot \log_2 \frac{\text{-}(x-1)}{\text{-}(x+1)}$

    $\displaystyle \displaystyle f(\text{-}x)\;=\;\left(\text{-}x-\log_3 \frac{2-x}{2+x}\right) \cdot \log_2 \frac{x-1}{x+1}$

    You're doing great!

    $\displaystyle \displaystyle f(\text{-}x) \;=\;\left(\text{-}x + (\text{-}1)\log_3\frac{2-x}{2+x}\right) \cdot \log_2\frac{x-1}{x+1} $

    $\displaystyle \displaystyle f(\text{-}x) \;=\;\left(\text{-}x + \log_3\left[\frac{2-x}{2+x}\right]^{-1}\right)\cdot\log_2\frac{x-1}{x+1}$

    $\displaystyle \displaystyle f(\text{-}x) \;=\;\left(\text{-}x + \log_3\frac{2+x}{2-x}\right) \cdit \log_2\frac{x-1}{x+1}$


    $\displaystyle \displaystyle f(\text{-}x) \;=\;(\text{-}1)\left(x - \log_3\frac{2+x}{2-x}\right)\cdot\log_2\frac{x-1}{x+1}$

    $\displaystyle \displaystyle f(\text{-}x) \;=\;\left(x - \log_3\frac{2+x}{2-x}\right)\cdot (\text{-}1)\cdot\log_2\frac{x-1}{x+1} $

    $\displaystyle \displaystyle f(\text{-}x) \;=\;\left(x - \log_3\frac{2+x}{2-x}\right)\cdot \log_2\left(\frac{x-1}{x+1}\right)^{-1} $

    $\displaystyle \displaystyle f(\text-}x) \;=\;\left(x - \log_3\frac{2+x}{2-x}\right) \cdot \log_2\frac{x+1}{x-1} \;=\;f(x)$

    . . . . . $\displaystyle \text{ta-}DAA!$

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  4. #4
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    interesting, thanks
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