Log problem

**livmed** · February 6th 2011, 04:47 AM

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

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

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

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

???

**VonNemo19** · February 6th 2011, 04:59 AM

Originally Posted by livmed

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

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

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

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

???

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 $\log{a}-\log{b}=\log\frac{a}{b}$ and the other properties that may be needed.

**Soroban** · February 6th 2011, 05:33 AM

Hello, livmed!

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

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

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

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

You're doing great!

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

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

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

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

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

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

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

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

**livmed** · February 6th 2011, 07:27 AM

interesting, thanks

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