1953 AHSME Problems/Problem 47

Problem

If $x>0$ , then the correct relationship is:

$\textbf{(A)}\ \log (1+x) = \frac{x}{1+x} \qquad \textbf{(B)}\ \log (1+x) < \frac{x}{1+x} \\ \textbf{(C)}\ \log(1+x) > x\qquad \textbf{(D)}\ \log (1+x) < x\qquad \textbf{(E)}\ \text{none of these}$

Solution

1953 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 46	Followed by Problem 48
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1953 AHSME Problems/Problem 47

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Solution

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