If e^{f(x)}=frac{10+x}{10-x}, x in(-10,10) an | vedclass

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(A) Given $e^{f(x)}=\frac{10+x}{10-x}$.Taking the natural logarithm on both sides,we get $f(x)=\log \left(\frac{10+x}{10-x}\right)$.We are given the r

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$0.5$

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(A) Given $e^{f(x)}=\frac{10+x}{10-x}$.Taking the natural logarithm on both sides,we get $f(x)=\log \left(\frac{10+x}{10-x}\right)$.We are given the relation $f(x)=k f\left(\frac{200 x}{100+x^2}\right)$.Substituting the expression for $f(x)$ on both sides:$\log \left(\frac{10+x}{10-x}\right) = k \log \left(\frac{10+\frac{200 x}{100+x^2}}{10-\frac{200 x}{100+x^2}}\right)$.Simplifying the argument of the logarithm on the right side:$\frac{10+\frac{200 x}{100+x^2}}{10-\frac{200 x}{100+x^2}} = \frac{10(100+x^2)+200x}{10(100+x^2)-200x} = \frac{1000+10x^2+200x}{1000+10x^2-200x} = \frac{10(x^2+20x+100)}{10(x^2-20x+100)} = \frac{(x+10)^2}{(10-x)^2}$.Thus,the equation becomes:$\log \left(\frac{10+x}{10-x}\right) = k \log \left(\frac{(x+10)^2}{(10-x)^2}\right)$.Using the property $\log(a^n) = n \log a$:$\log \left(\frac{10+x}{10-x}\right) = 2k \log \left(\frac{10+x}{10-x}\right)$.Comparing both sides,we get $2k = 1$,which implies $k = 0.5$.

If $e^{f(x)}=\frac{10+x}{10-x}, x \in(-10,10)$ and $f(x)=k f\left(\frac{200 x}{100+x^2}\right)$,then $k$ is equal to

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