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Difference between revisions of "1979 AHSME Problems/Problem 18"

(Created page with "== Problem 18 == To the nearest thousandth, <math>\log_{10}2</math> is <math>.301</math> and <math>\log_{10}3</math> is <math>.477</math>. Which of the following is the bes...")
 
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Solution by e_power_pi_times_i
 
Solution by e_power_pi_times_i
  
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Notice that <math>\log_5 10 = \log_5 2 + 1 = \frac{1}{\log_2 5} + 1</math>. So we are trying to find <math>\log_2 5</math>. Denote <math>\log_{10}2</math> as <math>x</math>. Then <math>\dfrac{1]{x} = \log_2 10 = \log_2 5 + 1</math>. Therefore <math>\log_2 5 = \frac{1}{x}-1</math>, and plugging this in gives <math>\log_5 10 = \frac{1}{\frac{1}{x}-1}+1 = \frac{x}{1-x}+1 = \frac{1}{1-x}</math>. Since <math>x</math> is around <math>\frac{3}{10}</math>, we substitute and get <math>\boxed{\textbf{(C) }\frac{10}{7}}</math>.
+
Notice that <math>\log_5 10 = \log_5 2 + 1 = \frac{1}{\log_2 5} + 1</math>. So we are trying to find <math>\log_2 5</math>. Denote <math>\log_{10}2</math> as <math>x</math>. Then <math>\frac{1}{x} = \log_2 10 = \log_2 5 + 1</math>. Therefore <math>\log_2 5 = \frac{1}{x}-1</math>, and plugging this in gives <math>\log_5 10 = \frac{1}{\frac{1}{x}-1}+1 = \frac{x}{1-x}+1 = \frac{1}{1-x}</math>. Since <math>x</math> is around <math>\frac{3}{10}</math>, we substitute and get <math>\boxed{\textbf{(C) }\frac{10}{7}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 15:07, 8 January 2017

Problem 18

To the nearest thousandth, $\log_{10}2$ is $.301$ and $\log_{10}3$ is $.477$. Which of the following is the best approximation of $\log_5 10$?

$\textbf{(A) }\frac{8}{7}\qquad \textbf{(B) }\frac{9}{7}\qquad \textbf{(C) }\frac{10}{7}\qquad \textbf{(D) }\frac{11}{7}\qquad \textbf{(E) }\frac{12}{7}$

Solution

Solution by e_power_pi_times_i

Notice that $\log_5 10 = \log_5 2 + 1 = \frac{1}{\log_2 5} + 1$. So we are trying to find $\log_2 5$. Denote $\log_{10}2$ as $x$. Then $\frac{1}{x} = \log_2 10 = \log_2 5 + 1$. Therefore $\log_2 5 = \frac{1}{x}-1$, and plugging this in gives $\log_5 10 = \frac{1}{\frac{1}{x}-1}+1 = \frac{x}{1-x}+1 = \frac{1}{1-x}$. Since $x$ is around $\frac{3}{10}$, we substitute and get $\boxed{\textbf{(C) }\frac{10}{7}}$.

See also

1979 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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All AHSME Problems and Solutions

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