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 "1966 AHSME Problems/Problem 12"

(→‎Solution)
m (→‎Solution)
 
(One intermediate revision by one other user not shown)
Line 6: Line 6:
 
== Solution ==
 
== Solution ==
 
We know that
 
We know that
−
<math>2^{6x+3}*4^{3x+6}=2^{6x+3}*(2^2)^{3x+6}=2^{6x+3}*2^{6x+12}=2^{12x+15}</math>.
+
<math>2^{6x+3}\cdot4^{3x+6}=2^{6x+3}\cdot(2^2)^{3x+6}=2^{6x+3}\cdot2^{6x+12}=2^{12x+15}</math>.
 
We also know that
 
We also know that
 
<math>8^{4x+5}=(2^3)^{4x+5}=2^{12x+15}</math>.
 
<math>8^{4x+5}=(2^3)^{4x+5}=2^{12x+15}</math>.
−
Thus there are infinite solutions to this equation since <math>2^{12x+15}</math> is always equal to <math>2^{12x+15}</math>. So the answer is <math>\text{greater than}</math> <math>3</math> <math>(E)</math>.
+
There are infinite solutions to the equation <math>2^{12x+15}=2^{12x+15}</math>, so the answer is <math>\boxed{\text{(E)  greater than 3}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 19:48, 24 March 2023

Problem

The number of real values of $x$ that satisfy the equation \[(2^{6x+3})(4^{3x+6})=8^{4x+5}\] is:

$\text{(A) zero} \qquad \text{(B) one} \qquad \text{(C) two} \qquad \text{(D) three} \qquad \text{(E) greater than 3}$

Solution

We know that $2^{6x+3}\cdot4^{3x+6}=2^{6x+3}\cdot(2^2)^{3x+6}=2^{6x+3}\cdot2^{6x+12}=2^{12x+15}$. We also know that $8^{4x+5}=(2^3)^{4x+5}=2^{12x+15}$. There are infinite solutions to the equation $2^{12x+15}=2^{12x+15}$, so the answer is $\boxed{\text{(E) greater than 3}}$.

See also

1966 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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

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=1966_AHSME_Problems/Problem_12&oldid=191288"