# 1963 AHSME Problems/Problem 30

## Problem

Let $F=\log\dfrac{1+x}{1-x}$. Find a new function $G$ by replacing each $x$ in $F$ by $\dfrac{3x+x^3}{1+3x^2}$, and simplify. The simplified expression $G$ is equal to: $\textbf{(A)}\ -F \qquad \textbf{(B)}\ F\qquad \textbf{(C)}\ 3F \qquad \textbf{(D)}\ F^3 \qquad \textbf{(E)}\ F^3-F$

## Solution

Replace the $x$ in $F$ with $\frac{3x+x^3}{1+3x^2}$. $$\log\dfrac{1 + \frac{3x+x^3}{1+3x^2} }{1 - \frac{3x+x^3}{1+3x^2} }$$ $$\log\dfrac{\frac{1+3x^2}{1+3x^2} + \frac{3x+x^3}{1+3x^2} }{\frac{1+3x^2}{1+3x^2} - \frac{3x+x^3}{1+3x^2} }$$ $$\log (\frac{1+3x+3x^2+x^3}{1+3x^2} \div \frac{1-3x+3x^2-x^3}{1+3x^2})$$ Recall that from the Binomial Theorem, $(1+x)^3 = 1+3x+3x^2+x^3$ and $(1-x)^3 = 1-3x+3x^2-x^3$. $$\log (\frac{(1+x)^3}{1+3x^2} \div \frac{(1-x)^3}{1+3x^2})$$ $$\log\dfrac{(1+x)^3}{(1-x)^3}$$ $$3 \log\dfrac{1+x}{1-x}$$ Thus, the expression is equivalent to $3F$, which is answer choice $\boxed{\textbf{(C)}}$.

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