Difference between revisions of "1967 AHSME Problems/Problem 26"

Latest revision as of 00:40, 16 August 2023

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 $\fbox{C}$

1967 AHSC (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 • 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.

@@ Line 29: / Line 29: @@
 == See also ==
-{{AHSME box|year=1967|num-b=25|num-a=27}}
+{{AHSME 40p box|year=1967|num-b=25|num-a=27}}
 [[Category:Introductory Algebra Problems]]
 {{MAA Notice}}

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Difference between revisions of "1967 AHSME Problems/Problem 26"

Latest revision as of 00:40, 16 August 2023

Problem

Solution

See also