Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1951 AHSME Problems/Problem 22"
m
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
The values of <math> a</math> in the equation:  <math> \log_{10}(a^2 \minus{} 15a) \equal{} 2</math> are:
+
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, \minus{} 5 \qquad\textbf{(C)}\ \frac {15 \pm \sqrt {305}}{2}</math>
+
<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>
  
Line 8: Line 8:
 
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, \minus{} 5}</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, - 5}</math>
  
 
== See Also ==
 
== See Also ==
Line 14: Line 14:
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1951_AHSME_Problems/Problem_22&oldid=68938"