Difference between revisions of "1965 AHSME Problems/Problem 21"
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== Solution == | == Solution == | ||
| − | <math>\fbox{D}</math> | + | |
| + | By the rules of [[logarithms]], <math>\log_{10}(x^2+3)-2\log_{10} x=\log_{10}(\frac{x^2+3}{x^2})=\log_{10}(1+\frac{3}{x^2})</math>. As <math>x</math> goes to infinity, <math>1+\frac{3}{x^2}</math> gets arbitrarily close to <math>1</math> (without ever reaching it), so <math>\log_{10}(1+\frac{3}{x^2})</math> gets arbitrarily close to <math>\log_{10}(1)=0</math> (without ever reaching it). Thus, we can choose a real <math>x>\frac{2}{3}</math> such that the given expression is <math>\fbox{\textbf{(D) }smaller than any positive number that might be specified}</math>. | ||
== See Also == | == See Also == | ||
| − | {{AHSME 40p box|year=1965|num-b= | + | {{AHSME 40p box|year=1965|num-b=20|num-a=22}} |
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
Latest revision as of 17:30, 18 July 2024
Problem 21
It is possible to choose 
in such a way that the value of 
is
Solution
By the rules of logarithms, 
. As 
goes to infinity, 
gets arbitrarily close to 
(without ever reaching it), so 
gets arbitrarily close to 
(without ever reaching it). Thus, we can choose a real 
such that the given expression is 
.
See Also
| 1965 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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. 
