Difference between revisions of "1951 AHSME Problems/Problem 22"
(Created page with "== Problem == The values of <math> a</math> in the equation: <math> \log_{10}(a^2 \minus{} 15a) \equal{} 2</math> are: <math> \textbf{(A)}\ \frac {15\pm\sqrt {233}}{2} \qquad\t...") |
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== Solution == | == Solution == | ||
| − | {{ | + | 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, \minus{} 5}</math> | ||
== See Also == | == See Also == | ||
Revision as of 20:35, 10 April 2013
Problem
The values of 
in the equation: $\log_{10}(a^2 \minus{} 15a) \equal{} 2$ (Error compiling LaTeX. Unknown error_msg) are:
$\textbf{(A)}\ \frac {15\pm\sqrt {233}}{2} \qquad\textbf{(B)}\ 20, \minus{} 5 \qquad\textbf{(C)}\ \frac {15 \pm \sqrt {305}}{2}$ (Error compiling LaTeX. Unknown error_msg)
Solution
Putting into exponential form, we get that 
Now we use the quadratic formula to solve for , and we get $a=\frac{15\pm\sqrt{625}}{2}\implies a=\boxed{\textbf{(B)}\ 20, \minus{} 5}$ (Error compiling LaTeX. Unknown error_msg)
See Also
| 1951 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 | ||
