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 "1968 AHSME Problems/Problem 2"

(Created page with "== Problem == The real value of <math>x</math> such that <math>64^{x-1}</math> divided by <math>4^{x-1}</math> equals <math>256^{2x}</math> is: <math>\text{(A) } -\frac{2}{3}\q...")
 
(Added solution to the problem)
 
(3 intermediate revisions by 2 users not shown)
Line 9: Line 9:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{}</math>
+
 
 +
Because <math>64=4^3</math> and <math>256=4^4</math>, we can change all of the numbers in the equation to exponents with base <math>4</math> and solve the equation:
 +
\begin{align*}
 +
\frac{64^{x-1}}{4^{x-1}}=256^{2x} \\
 +
\frac{(4^3)^{x-1}}{4^{x-1}}=(4^4)^{2x} \\
 +
\frac{4^{3x-3}}{4^{x-1}}=4^{8x} \\
 +
4^{2x-2}=4^{8x} \\
 +
2x-2=8x \\
 +
x-1=4x \\
 +
3x=-1 \\
 +
x=\frac{-1}{3} \\
 +
\end{align*}
 +
 
 +
Thus, our desired answer is <math>\fbox{(B) -1/3}</math>.
  
 
== See also ==
 
== See also ==
−
{{AHSME box|year=1968|num-b=1|num-a=2}}   
+
{{AHSME 35p box|year=1968|num-b=1|num-a=3}}   
  
 
[[Category: Introductory Algebra Problems]]
 
[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:30, 17 July 2024

Problem

The real value of $x$ such that $64^{x-1}$ divided by $4^{x-1}$ equals $256^{2x}$ is:

$\text{(A) } -\frac{2}{3}\quad\text{(B) } -\frac{1}{3}\quad\text{(C) } 0\quad\text{(D) } \frac{1}{4}\quad\text{(E) } \frac{3}{8}$



Solution

Because $64=4^3$ and $256=4^4$, we can change all of the numbers in the equation to exponents with base $4$ and solve the equation: \begin{align*} \frac{64^{x-1}}{4^{x-1}}=256^{2x} \\ \frac{(4^3)^{x-1}}{4^{x-1}}=(4^4)^{2x} \\ \frac{4^{3x-3}}{4^{x-1}}=4^{8x} \\ 4^{2x-2}=4^{8x} \\ 2x-2=8x \\ x-1=4x \\ 3x=-1 \\ x=\frac{-1}{3} \\ \end{align*}

Thus, our desired answer is $\fbox{(B) -1/3}$.

See also

1968 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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
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=1968_AHSME_Problems/Problem_2&oldid=222897"