1967 AHSME Problems/Problem 26

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 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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