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1966 AHSME Problems/Problem 9

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Problem

If $x = (\log_82)^{(\log_28)}$ , then $\log_3x$ equals:

$\text{(A)} \ - 3 \qquad \text{(B)} \ - \frac13 \qquad \text{(C)} \ \frac13 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 9$

Solution

By definition, $\log_8 2 = \frac 13$ and $\log_2 8 = 3$ , so $\log_3 \left(\frac{1}{3}\right)^3 = \log_3 3^{-3} = -3 \Rightarrow \mathrm{(A)}$ .

See also

1966 AHSME (Problems)
Preceded by Problem 8	Followed by Problem 10
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

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1966_AHSME_Problems/Problem_9"

Category: Introductory Algebra Problems

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