Difference between revisions of "1951 AHSME Problems/Problem 22"
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== Problem == | == Problem == | ||
− | The values of <math> a</math> in the equation: <math> \log_{10}(a^2 | + | The values of <math> a</math> in the equation: <math> \log_{10}(a^2 - 15a) = 2</math> are: |
− | <math> \textbf{(A)}\ \frac {15\pm\sqrt {233}}{2} \qquad\textbf{(B)}\ 20, | + | <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> | ||
− | 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, | + | 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 23:01, 13 March 2015
Problem
The values of in the equation: are:
Solution
Putting into exponential form, we get that
Now we use the quadratic formula to solve for , and we get
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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