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Difference between revisions of "1965 AHSME Problems/Problem 21"

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== Solution ==
 
== Solution ==
  
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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>.
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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 10:22, 29 July 2024

Problem 21

It is possible to choose $x > \frac {2}{3}$ in such a way that the value of $\log_{10}(x^2 + 3) - 2 \log_{10}x$ is

$\textbf{(A)}\ \text{negative} \qquad \textbf{(B) }\ \text{zero} \qquad \textbf{(C) }\ \text{one} \\ \textbf{(D) }\ \text{smaller than any positive number that might be specified} \\ \textbf{(E) }\ \text{greater than any positive number that might be specified}$

Solution

By the rules of logarithms, $\log_{10}(x^2+3)-2\log_{10} x=\log_{10}(\frac{x^2+3}{x^2})=\log_{10}(1+\frac{3}{x^2})$. As $x$ goes to infinity, $1+\frac{3}{x^2}$ gets arbitrarily close to $1$ (without ever reaching it), so $\log_{10}(1+\frac{3}{x^2})$ gets arbitrarily close to $\log_{10}(1)=0$ (without ever reaching it). Furthermore, because $1+\frac3{x^2} > 1$, $\log(1+\frac3{x^2})$ is never negative. Thus, we can choose a real $x>\frac{2}{3}$ such that the given expression is $\fbox{\textbf{(D) }smaller than any positive number that might be specified}$.

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

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