1963 AMC 12 Problem 30

Attempt Problem 30 of the 1963 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1963 AMC 12 solutions, or check the answer key.

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30.

Let
F=log⁡1+x1−x. F=\log\frac{1+x}{1-x}.
Form a new function GG by replacing each xx in FF by
3x+x31+3x2, \frac{3x+x^3}{1+3x^2},
and simplify. The simplified expression GG is equal to:
−F-F
FF
3F3F
F3F^3
F3−FF^3-F
Answer: C
Concepts:logarithmfactoringalgebraic manipulation
Small hint:
Call the substituted fraction uu and simplify 1+u1−u\frac{1+u}{1-u}
Big hint:
Its numerator and denominator factor as cubes
Solution:
For u=3x+x31+3x2,u=\frac{3x+x^3}{1+3x^2}, 1+u1−u=1+3x+3x2+x31−3x+3x2−x3=(1+x1−x)3. \begin{aligned} \frac{1+u}{1-u} &=\frac{1+3x+3x^2+x^3} {1-3x+3x^2-x^3}\\ &=\left(\frac{1+x}{1-x}\right)^3. \end{aligned} Therefore G=log⁡(1+x1−x)3,G=\log(\frac{1+x}{1-x})^3, so G=3F.G=3F.
Thus, the correct answer is C.
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