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 $\boxed{\textbf{(C)}}$.

See Also

1963 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 29
Followed by
Problem 31
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