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Difference between revisions of "1953 AHSME Problems/Problem 22"

(Created page with "This essentially simplifies to log base 3 of (3^3)(3^1/2)(3^2/3) Simplifying the equation, we get log base 3 of 3^3+1/2+2/3 Further simplifying, we get log base 3 of 3^25/6...")
 
m (I added a space.)
 
(One intermediate revision by the same user not shown)
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This essentially simplifies to log base 3 of (3^3)(3^1/2)(3^2/3)
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==Problem==
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The logarithm of <math>27\sqrt[4]{9}\sqrt[3]{9}</math> to the base <math>3</math> is:
  
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Simplifying the equation, we get log base 3 of 3^3+1/2+2/3
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<math>\textbf{(A)}\ 8\frac{1}{2} \qquad
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\textbf{(B)}\ 4\frac{1}{6} \qquad
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\textbf{(C)}\ 5 \qquad
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\textbf{(D)}\ 3 \qquad
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\textbf{(E)}\ \text{none of these}  </math>
  
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Further simplifying, we get log base 3 of 3^25/6
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==Solution==
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<math>27\sqrt[4]{9}\sqrt[3]{9}</math> can be rewritten as <math>3^3\cdot 3^\frac{1}{2}\cdot 3^\frac{2}{3}</math>. Using exponent rules, this simplifies to <math>3^\frac{25}{6}</math>.
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The problem wants us to find <math>\log_3{27\sqrt[4]{9}\sqrt[3]{9}}</math>. We just found that this is equal to <math>\log_3{3^\frac{25}{6}}</math>. Using logarithm rules, this is equal to <math>\frac{25}{6}\log_3{3}</math>, which is simply <math>\frac{25}{6}</math>. The answer is <math>\boxed{\text{(B)}\ 4\frac{1}{6}}</math>.
  
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We simplify this into a mixed number, and we obtain 4 and 1/6.
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==See Also==
  
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This is choice B.
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{{AHSME 50p box|year=1953|num-b=21|num-a=23}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 19:30, 27 November 2019

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 (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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