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Difference between revisions of "1951 AHSME Problems/Problem 22"
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== Problem ==
 
== Problem ==
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The values of <math> a</math> in the equation:  <math> \log_{10}(a^2 \minus{} 15a) \equal{} 2</math> are:
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The values of <math> a</math> in the equation:  <math> \log_{10}(a^2 - 15a) = 2</math> are:
  
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<math> \textbf{(A)}\ \frac {15\pm\sqrt {233}}{2} \qquad\textbf{(B)}\ 20, \minus{} 5 \qquad\textbf{(C)}\ \frac {15 \pm \sqrt {305}}{2}</math>
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<math> \textbf{(A)}\ \frac {15\pm\sqrt {233}}{2} \qquad\textbf{(B)}\ 20, - 5 \qquad\textbf{(C)}\ \frac {15 \pm \sqrt {305}}{2}</math>
 
<math> \textbf{(D)}\ \pm20 \qquad\textbf{(E)}\ \text{none of these}</math>
 
<math> \textbf{(D)}\ \pm20 \qquad\textbf{(E)}\ \text{none of these}</math>
  
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Putting into exponential form, we get that <math>10^2=a^2-15a\Rightarrow a^2-15a-100=0</math>
 
Putting into exponential form, we get that <math>10^2=a^2-15a\Rightarrow a^2-15a-100=0</math>
  
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Now we use the quadratic formula to solve for <math>a</math>, and we get <math>a=\frac{15\pm\sqrt{625}}{2}\implies a=\boxed{\textbf{(B)}\ 20, \minus{} 5}</math>
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Now we use the quadratic formula to solve for <math>a</math>, and we get <math>a=\frac{15\pm\sqrt{625}}{2}\implies a=\boxed{\textbf{(B)}\ 20, - 5}</math>
  
 
== See Also ==
 
== See Also ==

Latest revision as of 22:01, 13 March 2015

Problem

The values of $a$ in the equation: $\log_{10}(a^2 - 15a) = 2$ are:

$\textbf{(A)}\ \frac {15\pm\sqrt {233}}{2} \qquad\textbf{(B)}\ 20, - 5 \qquad\textbf{(C)}\ \frac {15 \pm \sqrt {305}}{2}$ $\textbf{(D)}\ \pm20 \qquad\textbf{(E)}\ \text{none of these}$

Solution

Putting into exponential form, we get that $10^2=a^2-15a\Rightarrow a^2-15a-100=0$

Now we use the quadratic formula to solve for $a$, and we get $a=\frac{15\pm\sqrt{625}}{2}\implies a=\boxed{\textbf{(B)}\ 20, - 5}$

See Also

1951 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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All AHSME Problems and Solutions

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