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Math question help (Functions) (1 Viewer)

Thread starter Accurate
Start date Sep 1, 2019

A

Accurate

clix

#1

How do you do this ..

D

Drongoski

Well-Known Member

#2

$g(\frac {2x}{1+x^2}) =log\left ( \frac {1+\frac {2x}{1+x^2}}{1- \frac {2x}{1+x^2}} \right) = log (\frac {1+x^2+2x}{1+x^2-2x}) =log (\frac {1+x}{1-x})^2 \\ \\ = 2log(\frac {1+x}{1-x}) = 2g(x)$

Last edited: Sep 1, 2019

A

Accurate

clix

#3

Drongoski said:
$g(\frac {2x}{1+x^2}) =log\left ( \frac {1+\frac {2x}{1+x^2}}{1- \frac {2x}{1+x^2}} \right) = log (\frac {1+x^2+2x}{1+x^2-2x}) =log (\frac {1+x}{1-x})^2 \\ \\ = 2log(\frac {1+x}{1-x}) = 2g(x)$

Thank you.

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