1953 AHSME Problems/Problem 22

Revision as of 19:22, 27 November 2019 by Anmol04 (talk | contribs) (The solution was originally just the solution, so I added everything else and rewrote the solution.)

Problem

The logarithm of $27\sqrt[4]{9}\sqrt[3]{9}$ to the base $3$ is:

$\textbf{(A)}\ 8\frac{1}{2} \qquad \textbf{(B)}\ 4\frac{1}{6} \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ \text{none of these}$

Solution

$27\sqrt[4]{9}\sqrt[3]{9}$ can be rewritten as $3^3\cdot 3^\frac{1}{2}\cdot 3^\frac{2}{3}$. Using exponent rules, this simplifies to $3^\frac{25}{6}$. The problem wants us to find $\log_3{27\sqrt[4]{9}\sqrt[3]{9}}$. We just found that this is equal to $\log_3{3^\frac{25}{6}}$. Using logarithm rules, this is equal to $\frac{25}{6}\log_3{3}$, which is simply $\frac{25}{6}$. The answer is $\boxed{\text{(B)}\ 4\frac{1}{6}}$.

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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