Are Trigonometric Substitutions Needed?

Algebra Level 4

Let f ( x ) = log ⁡ ( 1 + x 1 − x ) f(x) = \log \left( \dfrac{1+x}{1-x} \right) , where − 1 < x < 1 -1<x<1 . Which of the following choices is equal to f ( 3 x + x 3 1 + 3 x 2 ) − f ( 2 x 1 + x 2 ) f \left( \dfrac{3x+x^3}{1+3x^2} \right) - f\left( \dfrac{2x}{1+x^2}\right) ?

[ f ( x ) ] 3 [f(x)]^3 − f ( x ) -f(x) [ f ( x ) ] 2 [f(x)]^2 f ( x ) f(x)

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Sabhrant Sachan by Sabhrant Sachan , 21, India

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1 solution

Chew-Seong Cheong
Apr 17, 2016

Relevant wiki: Logarithmic Functions - Problem Solving - Hard

f ( 3 x + x 3 1 + 3 x ) − f ( 2 x 1 + x 2 ) = log ⁡ ( 1 + 3 x + x 3 1 + 3 x 1 − 3 x + x 3 1 + 3 x ) − log ⁡ ( 1 + 2 x 1 + x 2 1 − 2 x 1 + x 2 ) = log ⁡ ( 1 + 3 x 2 + 3 x + x 3 1 + 3 x 2 − 3 x − x 3 ) − log ⁡ ( 1 + x 2 + 2 x 1 + x 2 − 2 x ) = log ⁡ [ ( 1 + x 1 − x ) 3 ] − log ⁡ [ ( 1 + x 1 − x ) 2 ] = 3 log ⁡ ( 1 + x 1 − x ) − 2 log ⁡ ( 1 + x 1 − x ) = log ⁡ ( 1 + x 1 − x ) = f ( x ) \begin{aligned} f \left(\frac{3x+x^3}{1+3x} \right) - f \left(\frac{2x}{1+x^2} \right) & = \log \left( \frac {1 + \frac{3x+x^3}{1+3x}} {1 - \frac{3x+x^3}{1+3x}} \right) - \log \left( \frac {1 + \frac{2x}{1+x^2}} {1 - \frac{2x}{1+x^2}} \right) \\ & = \log \left(\frac{1+3x^2+3x+x^3}{1+3x^2-3x-x^3} \right) - \log \left(\frac{1+x^2+2x}{1+x^2-2x} \right) \\ & = \log \left[ \left( \frac{1+x}{1-x} \right)^3 \right] - \log \left[ \left( \frac{1+x}{1-x} \right)^2 \right] \\ & = 3 \log \left( \frac{1+x}{1-x} \right) - 2 \log \left( \frac{1+x}{1-x} \right) \\ & = \log \left( \frac{1+x}{1-x} \right) = \boxed{f(x)} \end{aligned}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

you got it right :)

Sabhrant Sachan - 5 years, 1 month ago

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