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 1(Cheap)== | ||
| + | Plug in <math>x=9</math>. Then, you can see that the answer is <math>\fbox{D}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AHSME 50p box|year=1953|num-b=46|num-a=48}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 17:51, 22 April 2020
Problem
If 
, then the correct relationship is:
Solution 1(Cheap)
Plug in 
. Then, you can see that the answer is 
.
See Also
| 1953 AHSC (Problems • Answer Key • Resources) | ||
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
