1953 AHSME Problems/Problem 47

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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 1(Cheap)

Plug in $x=9$. Then, you can see that the answer is $\fbox{D}$.

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 46
Followed by
Problem 48
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