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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>
  

Revision as of 18:23, 10 March 2017

Problem

If one uses only the tabular information $10^3=1000$, $10^4=10,000$, $2^{10}=1024$, $2^{11}=2048$, $2^{12}=4096$, $2^{13}=8192$, then the strongest statement one can make for $\log_{10}{2}$ is that it lies between:

$\textbf{(A)}\ \frac{3}{10} \; \text{and} \; \frac{4}{11}\qquad \textbf{(B)}\ \frac{3}{10} \; \text{and} \; \frac{4}{12}\qquad \textbf{(C)}\ \frac{3}{10} \; \text{and} \; \frac{4}{13}\qquad \textbf{(D)}\ \frac{3}{10} \; \text{and} \; \frac{40}{132}\qquad \textbf{(E)}\ \frac{3}{11} \; \text{and} \; \frac{40}{132}$

Solution

Since $1024$ is greater than $1000$.

$\log (1024) > 3$

$10 * \log (2) > 3$

and $\log (2) > 3/10$.


Similarly, $8192 < 10000$, so $\log (8192) < 4$

$13 * \log (2) < 4$

and $\log (2) < 4/13$


Therefore $3/10 < \log 2 < 4/13$ so the answer is $\fbox{C}$

See also

1967 AHSME (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
All AHSME Problems and Solutions

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