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Algebra Level 2

Function $f$ is defined as $f(x) = \ln(1+x) - \ln(1-x)$ , Then

$f \left(\frac{2x}{1+x^2} \right) = kf(x)$

where $k$ is an integer and $x$ does not equal to 0. Find the value of $k$ .

1 2 100 200

2 solutions

Chew-Seong Cheong
Dec 5, 2020

$\begin{aligned} f(x) & = \ln (1+x) - \ln (1-x) = \ln \left(\frac {1+x}{1-x} \right) \\ \implies f \left(\frac {2x}{1+x^2} \right) & = \ln \left(\frac {1+\frac {2x}{1+x^2}}{1-\frac {2x}{1+x^2}} \right) \\ & = \ln \left(\frac {x^2 + 2x + 1}{x^2-2x+1} \right) \\ & = \ln \left(\frac {1+x}{1-x} \right)^2 \\ & = 2 \ln \left(\frac {1+x}{1-x} \right) \\ & = 2 f(x) \end{aligned}$

Therefore $k = \boxed 2$ .

Ethan Mandelez
Dec 5, 2020

@Ethan Mandelez , you should learn up how to use LaTex for your solutions. Note that it should be \ln x $\ln x$ (the ln is not italic but x is italic and there is a space between ln and x) and not ln x $ln x$ (all are italic and no space in between). Can used display fraction \dfrac \pi 2 $\dfrac \pi 2$ instead of just \frac 12 $\frac 12$ . Use \left(\dfrac mn \right) $\left(\dfrac mn \right)$ , then the brackets size is automatically adjusted.

Chew-Seong Cheong - 6 months, 1 week ago

thanks for the tip! I'll do it next time 👍

Ethan Mandelez - 6 months, 1 week ago

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