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Difference between revisions of "1954 AHSME Problems/Problem 38"

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==Solution==
 
==Solution==
  
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Taking the logarithm in base <math>3</math> of both sides, we get <math>x+3 = \log_3 135</math>. Using the property <math>\log ab = \log a + \log b</math>, we get <math>x+3 = \log_3 5 + \log_3 3^3</math>, or <math>x = \log_3 5</math>. Converting into base <math>10</math> gives <math>x = \frac{\log 5}{\log 3} = \frac{1 - \log 2}{\log 3}</math>. Now, plugging in the values yeilds <math>\boxed{\textbf{(C) \ } 1.67 }</math>.
+
Taking the logarithm in base <math>3</math> of both sides, we get <math>x+3 = \log_3 135</math>. Using the property <math>\log ab = \log a + \log b</math>, we get <math>x+3 = \log_3 5 + \log_3 3^3</math>, or <math>x = \log_3 5</math>. Converting into base <math>10</math> gives <math>x = \frac{\log 5}{\log 3} = \frac{1 - \log 2}{\log 3}</math>. Now, plugging in the values yeilds <math>\boxed{\textbf{(B) \ } 1.47  }</math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME 50p box|year=1954|num-b=37|num-a=39}}
 
{{AHSME 50p box|year=1954|num-b=37|num-a=39}}

Revision as of 11:43, 6 August 2012

Problem

If $\log 2 = .3010$ and $\log 3 = .4771$, the value of $x$ when $3^{x+3} = 135$ is approximately

$\textbf{(A) \ } 5 \qquad \textbf{(B) \ } 1.47 \qquad \textbf{(C) \ } 1.67 \qquad \textbf{(D) \ } 1.78 \qquad \textbf{(E) \ } 1.63$

Solution

Taking the logarithm in base $3$ of both sides, we get $x+3 = \log_3 135$. Using the property $\log ab = \log a + \log b$, we get $x+3 = \log_3 5 + \log_3 3^3$, or $x = \log_3 5$. Converting into base $10$ gives $x = \frac{\log 5}{\log 3} = \frac{1 - \log 2}{\log 3}$. Now, plugging in the values yeilds $\boxed{\textbf{(B) \ } 1.47 }$.

See Also

1954 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 37 		Followed by
Problem 39
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All AHSME Problems and Solutions
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