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 "2009 AMC 12B Problems/Problem 15"

(New page: == Problem == Assume <math>0 < r < 3</math>. Below are five equations for <math>x</math>. Which equation has the largest solution <math>x</math>? <math>\textbf{(A)}\ 3(1 + r)^x = 7\qqua...)
 
(→‎Solution)
 
(5 intermediate revisions by 4 users not shown)
Line 5: Line 5:
 
<math>\textbf{(D)}\ 3(1 + \sqrt {r})^x = 7\qquad \textbf{(E)}\ 3(1 + 1/r)^x = 7</math>
 
<math>\textbf{(D)}\ 3(1 + \sqrt {r})^x = 7\qquad \textbf{(E)}\ 3(1 + 1/r)^x = 7</math>
  
−
== Solution ==
+
== Solution 1 ==
  
−
Intuitivelly, <math>x</math> will be largest for that option for which the value in the parentheses is smallest.
+
'''(B)''' Intuitively, <math>x</math> will be largest for that option for which the value in the parentheses is smallest.
  
 
Formally, first note that each of the values in parentheses is larger than <math>1</math>. Now, each of the options is of the form <math>3f(r)^x = 7</math>. This can be rewritten as <math>x\log f(r) = \log\frac 73</math>. As <math>f(r)>1</math>, we have <math>\log f(r)>0</math>. Thus <math>x</math> is the largest for the option for which <math>\log f(r)</math> is smallest. And as <math>\log f(r)</math> is an increasing function, this is the option for which <math>f(r)</math> is smallest.
 
Formally, first note that each of the values in parentheses is larger than <math>1</math>. Now, each of the options is of the form <math>3f(r)^x = 7</math>. This can be rewritten as <math>x\log f(r) = \log\frac 73</math>. As <math>f(r)>1</math>, we have <math>\log f(r)>0</math>. Thus <math>x</math> is the largest for the option for which <math>\log f(r)</math> is smallest. And as <math>\log f(r)</math> is an increasing function, this is the option for which <math>f(r)</math> is smallest.
Line 19: Line 19:
 
And finally, <math>r/100 < 1</math>, therefore <math>r^2/100 < r</math>, and as both sides are positive, we can take the square root and get <math>r/10 < \sqrt r</math>.
 
And finally, <math>r/100 < 1</math>, therefore <math>r^2/100 < r</math>, and as both sides are positive, we can take the square root and get <math>r/10 < \sqrt r</math>.
  
−
Thus the answer is <math>\boxed{\text{(B) } 3(1 + r/10)^x = 7}</math>.
+
Thus the answer is <math>\boxed{\textbf{(B) } 3\left(1+\frac{r}{10}\right)^x = 7}</math>.
  
 +
== Solution 2 ==
 +
As stated in the first solution, <math>x</math> will be largest when the value in the parenthesis is smallest.
 +
 +
We can plug in a value for <math>r</math> (for instance, <math>r=2</math>) and see which parenthesis has the smallest value. Doing so, we see that B is the answer.
 +
 +
-Made_in_2004
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2009|ab=B|num-b=14|num-a=16}}
 
{{AMC12 box|year=2009|ab=B|num-b=14|num-a=16}}
 +
{{MAA Notice}}

Latest revision as of 15:58, 27 December 2020

Problem

Assume $0 < r < 3$. Below are five equations for $x$. Which equation has the largest solution $x$?

$\textbf{(A)}\ 3(1 + r)^x = 7\qquad \textbf{(B)}\ 3(1 + r/10)^x = 7\qquad \textbf{(C)}\ 3(1 + 2r)^x = 7$ $\textbf{(D)}\ 3(1 + \sqrt {r})^x = 7\qquad \textbf{(E)}\ 3(1 + 1/r)^x = 7$

Solution 1

(B) Intuitively, $x$ will be largest for that option for which the value in the parentheses is smallest.

Formally, first note that each of the values in parentheses is larger than $1$. Now, each of the options is of the form $3f(r)^x = 7$. This can be rewritten as $x\log f(r) = \log\frac 73$. As $f(r)>1$, we have $\log f(r)>0$. Thus $x$ is the largest for the option for which $\log f(r)$ is smallest. And as $\log f(r)$ is an increasing function, this is the option for which $f(r)$ is smallest.

We now get the following easier problem: Given that $0<r<3$, find the smallest value in the set $\{ 1+r, 1+r/10, 1+2r, 1+\sqrt r, 1+1/r\}$.

Clearly $1+r/10$ is smaller than the first and the third option.

We have $r^2 < 10$, dividing both sides by $10r$ we get $r/10 < 1/r$.

And finally, $r/100 < 1$, therefore $r^2/100 < r$, and as both sides are positive, we can take the square root and get $r/10 < \sqrt r$.

Thus the answer is $\boxed{\textbf{(B) } 3\left(1+\frac{r}{10}\right)^x = 7}$.

Solution 2

As stated in the first solution, $x$ will be largest when the value in the parenthesis is smallest.

We can plug in a value for $r$ (for instance, $r=2$) and see which parenthesis has the smallest value. Doing so, we see that B is the answer.

-Made_in_2004

See Also

2009 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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
All AMC 12 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=2009_AMC_12B_Problems/Problem_15&oldid=140761"