Difference between revisions of "1967 AHSME Problems/Problem 26"
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<math>\log (1024) > 3</math> | <math>\log (1024) > 3</math> | ||
| − | <math>10 * log (2) > 3</math> | + | <math>10 * \log (2) > 3</math> |
| − | and <math>log (2) > 3/10</math>. | + | and <math>\log (2) > 3/10</math>. |
| − | Similarly, <math>8192 < 10000</math>, so <math>log (8192) < 4</math> | + | Similarly, <math>8192 < 10000</math>, so <math>\log (8192) < 4</math> |
| − | <math>13 * log (2) < 4</math> | + | <math>13 * \log (2) < 4</math> |
| − | and <math>log (2) < 4/13</math> | + | and <math>\log (2) < 4/13</math> |
| − | Therefore <math>3/10 < log 2 < 4/13</math> | + | Therefore <math>3/10 < \log 2 < 4/13</math> |
so the answer is <math>\fbox{C}</math> | so the answer is <math>\fbox{C}</math> | ||
== See also == | == See also == | ||
| − | {{AHSME box|year=1967|num-b=25|num-a=27}} | + | {{AHSME 40p box|year=1967|num-b=25|num-a=27}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 00:40, 16 August 2023
Problem
If one uses only the tabular information 
, 
, 
, 
, 
, 
, then the strongest statement one can make for 
is that it lies between:
Solution
Since 
is greater than 
.
and 
.
Similarly, 
, so 
and 
Therefore 
so the answer is 
See also
| 1967 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 25 |
Followed by Problem 27 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
