Difference between revisions of "1965 AHSME Problems/Problem 21"
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== Solution == | == Solution == | ||
− | 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>. | + | 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). Furthermore, because <math>1+\frac3{x^2} > 1</math>, <math>\log(1+\frac3{x^2})</math> is never negative. 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 == |
Latest revision as of 11:22, 29 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). Furthermore, because
,
is never negative. 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 | |
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All AHSME Problems and Solutions |
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