Difference between revisions of "1953 AHSME Problems/Problem 47"
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− | + | ==Problem== | |
+ | |||
+ | If <math>x>0</math>, then the correct relationship is: | ||
+ | |||
+ | <math>\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} </math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AHSME 50p box|year=1953|num-b=46|num-a=48}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 23:40, 14 February 2020
Problem
If , then the correct relationship is:
Solution
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 46 |
Followed by Problem 48 | |
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All AHSME Problems and Solutions |
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