Difference between revisions of "1953 AHSME Problems/Problem 47"

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==Problem==
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If <math>x>0</math>, then the correct relationship is:
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<math>\textbf{(A)}\ \log (1+x) = \frac{x}{1+x} \qquad
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\textbf{(B)}\ \log (1+x) < \frac{x}{1+x} \\
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\textbf{(C)}\ \log(1+x) > x\qquad
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\textbf{(D)}\ \log (1+x) < x\qquad
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\textbf{(E)}\ \text{none of these}  </math>
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==Solution==
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==See Also==
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{{AHSME 50p box|year=1953|num-b=46|num-a=48}}
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{{MAA Notice}}

Latest revision as of 23:40, 14 February 2020

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

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 46
Followed by
Problem 48
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions


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