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
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

Difference between revisions of "1965 AHSME Problems/Problem 1"

(Created page with "== Problem== 1. The number of real values of <math>x</math> satisfying the equation <math>2^{2x^2 - 7x + 5} = 1</math> is: <math>\textbf{(A)}\ 0 \qquad \textbf{(B) }\ 1 \qq...")
 
m (added link)
 
(7 intermediate revisions by 4 users not shown)
Line 1: Line 1:
 
== Problem==
 
== Problem==
  
−
1. The number of real values of <math>x</math> satisfying the equation <math>2^{2x^2 - 7x + 5} = 1</math> is:
+
The number of real values of <math>x</math> satisfying the equation <math>2^{2x^2 - 7x + 5} = 1</math> is:
  
 
<math>\textbf{(A)}\ 0 \qquad  \textbf{(B) }\ 1 \qquad  \textbf{(C) }\ 2 \qquad  \textbf{(D) }\ 3 \qquad  \textbf{(E) }\ \text{more than 4}</math>
 
<math>\textbf{(A)}\ 0 \qquad  \textbf{(B) }\ 1 \qquad  \textbf{(C) }\ 2 \qquad  \textbf{(D) }\ 3 \qquad  \textbf{(E) }\ \text{more than 4}</math>
 +
 +
 +
==Solution 1==
 +
Solution by e_power_pi_times_i
 +
 +
Take the logarithm with a base of <math>2</math> to both sides, resulting in the equation <math>2x^2-7x+5 = 0</math>. Factoring results in <math>(2x-5)(x-1) = 0</math>, so there are <math>\boxed{\textbf{(C) } 2}</math> real solutions.
 +
 +
==Solution 2==
 +
Notice that <math>a^0=1, a>0</math>. So <math>2^0=1</math>. So <math>2x^2-7x+5=0</math>. Evaluating the [[discriminant]], we see that it is equal to <math>7^2-4*2*5=9</math>. So this means that the equation has two real solutions. Therefore, select <math>\fbox{C}</math>.
 +
 +
~hastapasta
 +
 +
==See Also==
 +
{{AHSME box|year=1965|before=First Question|num-a=2}}
 +
{{MAA Notice}}
 +
[[Category:AHSME]][[Category:AHSME Problems]]

Latest revision as of 08:10, 18 July 2024

Problem

The number of real values of $x$ satisfying the equation $2^{2x^2 - 7x + 5} = 1$ is:

$\textbf{(A)}\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 3 \qquad \textbf{(E) }\ \text{more than 4}$


Solution 1

Solution by e_power_pi_times_i

Take the logarithm with a base of $2$ to both sides, resulting in the equation $2x^2-7x+5 = 0$. Factoring results in $(2x-5)(x-1) = 0$, so there are $\boxed{\textbf{(C) } 2}$ real solutions.

Solution 2

Notice that $a^0=1, a>0$. So $2^0=1$. So $2x^2-7x+5=0$. Evaluating the discriminant, we see that it is equal to $7^2-4*2*5=9$. So this means that the equation has two real solutions. Therefore, select $\fbox{C}$.

~hastapasta

See Also

1965 AHSME (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
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
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1965_AHSME_Problems/Problem_1&oldid=222954"