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 "2008 AMC 12A Problems/Problem 16"

(→‎See Also)
m (→‎Solution 4 (mimimal manipulation))
 
(9 intermediate revisions by 4 users not shown)
Line 1: Line 1:
−
==Problem==  
+
== Problem ==  
 
The numbers <math>\log(a^3b^7)</math>, <math>\log(a^5b^{12})</math>, and <math>\log(a^8b^{15})</math> are the first three terms of an [[arithmetic sequence]], and the <math>12^\text{th}</math> term of the sequence is <math>\log{b^n}</math>. What is <math>n</math>?
 
The numbers <math>\log(a^3b^7)</math>, <math>\log(a^5b^{12})</math>, and <math>\log(a^8b^{15})</math> are the first three terms of an [[arithmetic sequence]], and the <math>12^\text{th}</math> term of the sequence is <math>\log{b^n}</math>. What is <math>n</math>?
  
 
<math>\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 56\qquad\mathrm{(C)}\ 76\qquad\mathrm{(D)}\ 112\qquad\mathrm{(E)}\ 143</math>
 
<math>\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 56\qquad\mathrm{(C)}\ 76\qquad\mathrm{(D)}\ 112\qquad\mathrm{(E)}\ 143</math>
  
−
__TOC__
+
== Solutions ==
−
==Solution==  
+
=== Solution 1 ===
−
===Solution 1===
 
 
Let <math>A = \log(a)</math> and <math>B = \log(b)</math>.  
 
Let <math>A = \log(a)</math> and <math>B = \log(b)</math>.  
  
Line 15: Line 14:
 
Since the first three terms in the sequence are <math>13B</math>, <math>22B</math>, and <math>31B</math>, the <math>k</math>th term is <math>(9k + 4)B</math>.
 
Since the first three terms in the sequence are <math>13B</math>, <math>22B</math>, and <math>31B</math>, the <math>k</math>th term is <math>(9k + 4)B</math>.
 
   
 
   
−
Thus the <math>12^\text{th}</math> term is <math>(9\cdot12 + 4)B = 112B = nB \Rightarrow n = 112\Rightarrow D</math>.  
+
Thus the <math>12^\text{th}</math> term is <math>(9\cdot12 + 4)B = 112B = nB \Rightarrow n = 112\Rightarrow \boxed{D}</math>.
  
−
===Solution 2===
+
=== Solution 2 ===
 
If <math>\log(a^3b^7)</math>, <math>\log(a^5b^{12})</math>, and <math>\log(a^8b^{15})</math> are in [[arithmetic progression]], then <math>a^3b^7</math>, <math>a^5b^{12}</math>, and <math>a^8b^{15}</math> are in [[geometric progression]]. Therefore,
 
If <math>\log(a^3b^7)</math>, <math>\log(a^5b^{12})</math>, and <math>\log(a^8b^{15})</math> are in [[arithmetic progression]], then <math>a^3b^7</math>, <math>a^5b^{12}</math>, and <math>a^8b^{15}</math> are in [[geometric progression]]. Therefore,
  
 
<cmath>a^2b^5=a^3b^3 \Rightarrow a=b^2</cmath>
 
<cmath>a^2b^5=a^3b^3 \Rightarrow a=b^2</cmath>
  
−
Therefore, <math>a^3b^7=b^{13}</math>, <math>a^5b^{12}=b^{22}</math>, therefore the 12th term in the sequence is <math>b^{13+9*11}=b^{112} \Rightarrow D</math>
+
Therefore, <math>a^3b^7=b^{13}</math>, <math>a^5b^{12}=b^{22}</math>, therefore the 12th term in the sequence is <math>b^{13+9*11}=b^{112} \Rightarrow \boxed{D}</math>
  
−
==See Also==  
+
=== Solution 3 ===
 +
<math>\ \text{If a, b, and c are in a arithmetic progression then } b = \frac{a+c}{2} \text{ which means}</math>
 +
<math>\ \log(a^5b^{12}) = \frac{\log(a^3b^7) + \log(a^8b^{15})}{2}= \frac{\log(a^{11}b^{22})}{2} \text{ therefore}</math>
 +
<math>\ 2\log(a^5b^{12}) = \log(a^{10}b^{24}) = \log(a^{11}b^{22}) \Rightarrow a=b^2</math>
 +
<math>\ \text{This means that the Kth term of the series would be } \log(b^{13+9(k-1)})</math>
 +
<math>\ \text{The 12th term would be } \log(b^{112})  \Rightarrow n=112  \Rightarrow D </math>
 +
 
 +
=== Solution 4 (mimimal manipulation)===
 +
Given the first three terms form an arithmetic progression, we have:
 +
<cmath>a = \log(a^3b^7)</cmath>
 +
<cmath>a+d = \log(a^5b^{12})</cmath>
 +
<cmath>a+2d = \log(a^8b^{15}).</cmath>
 +
Subtracting the first equation from the second and the third from the second, respectively, gives us these two expressions for <math>d</math>:
 +
<cmath>d = \log \left( \frac{a^5b^{12}}{a^3b^7} \right) = \log a^2b^5</cmath>
 +
<cmath>d = \log \left( \frac{a^8b^{15}}{a^5b^{12}} \right) = \log a^3b^3.</cmath>
 +
The desired <math>12</math>th term in the sequence is <math>a+11d</math>, so we can substitute our values for <math>a</math> and <math>d</math> (using either one of our two expressions for <math>d</math>):
 +
<cmath>a+11d = \log a^3b^7 + 11\log(a^2b^5)</cmath>
 +
<cmath> = \log a^3b^7 + \log(a^{22}b^{55})</cmath>
 +
<cmath> = \log a^{25}b^{62}.</cmath>
 +
The answer must be expressed as <math>\log(b^n)</math>, however. We're in luck: the two different yet equal expressions for <math>d</math> allow us to express <math>a</math> and <math>b</math> in terms of each other:
 +
<cmath>\log a^2b^5 = \log a^3b^3</cmath>
 +
<cmath>a^2b^5 = a^3b^3</cmath>
 +
<cmath>a=b^2.</cmath>
 +
Plugging in <math>a=b^2</math>, we have:
 +
<cmath>a+11d = \log b^{50}b^{62}</cmath>
 +
<cmath> = \log b^{112} \Rightarrow \boxed{D}.</cmath>
 +
~ Jingwei325 <math>\smiley</math>
 +
 
 +
== See Also ==  
 
{{AMC12 box|year=2008|ab=A|num-b=15|num-a=17}}
 
{{AMC12 box|year=2008|ab=A|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:33, 9 October 2022

Problem

The numbers $\log(a^3b^7)$, $\log(a^5b^{12})$, and $\log(a^8b^{15})$ are the first three terms of an arithmetic sequence, and the $12^\text{th}$ term of the sequence is $\log{b^n}$. What is $n$?

$\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 56\qquad\mathrm{(C)}\ 76\qquad\mathrm{(D)}\ 112\qquad\mathrm{(E)}\ 143$

Solutions

Solution 1

Let $A = \log(a)$ and $B = \log(b)$.

The first three terms of the arithmetic sequence are $3A + 7B$, $5A + 12B$, and $8A + 15B$, and the $12^\text{th}$ term is $nB$.

Thus, $2(5A + 12B) = (3A + 7B) + (8A + 15B) \Rightarrow A = 2B$.

Since the first three terms in the sequence are $13B$, $22B$, and $31B$, the $k$th term is $(9k + 4)B$.

Thus the $12^\text{th}$ term is $(9\cdot12 + 4)B = 112B = nB \Rightarrow n = 112\Rightarrow \boxed{D}$.

Solution 2

If $\log(a^3b^7)$, $\log(a^5b^{12})$, and $\log(a^8b^{15})$ are in arithmetic progression, then $a^3b^7$, $a^5b^{12}$, and $a^8b^{15}$ are in geometric progression. Therefore,

\[a^2b^5=a^3b^3 \Rightarrow a=b^2\]

Therefore, $a^3b^7=b^{13}$, $a^5b^{12}=b^{22}$, therefore the 12th term in the sequence is $b^{13+9*11}=b^{112} \Rightarrow \boxed{D}$

Solution 3

$\ \text{If a, b, and c are in a arithmetic progression then } b = \frac{a+c}{2} \text{ which means}$ $\ \log(a^5b^{12}) = \frac{\log(a^3b^7) + \log(a^8b^{15})}{2}= \frac{\log(a^{11}b^{22})}{2} \text{ therefore}$ $\ 2\log(a^5b^{12}) = \log(a^{10}b^{24}) = \log(a^{11}b^{22}) \Rightarrow a=b^2$ $\ \text{This means that the Kth term of the series would be } \log(b^{13+9(k-1)})$ $\ \text{The 12th term would be } \log(b^{112}) \Rightarrow n=112 \Rightarrow D$

Solution 4 (mimimal manipulation)

Given the first three terms form an arithmetic progression, we have: \[a = \log(a^3b^7)\] \[a+d = \log(a^5b^{12})\] \[a+2d = \log(a^8b^{15}).\] Subtracting the first equation from the second and the third from the second, respectively, gives us these two expressions for $d$: \[d = \log \left( \frac{a^5b^{12}}{a^3b^7} \right) = \log a^2b^5\] \[d = \log \left( \frac{a^8b^{15}}{a^5b^{12}} \right) = \log a^3b^3.\] The desired $12$th term in the sequence is $a+11d$, so we can substitute our values for $a$ and $d$ (using either one of our two expressions for $d$): \[a+11d = \log a^3b^7 + 11\log(a^2b^5)\] \[= \log a^3b^7 + \log(a^{22}b^{55})\] \[= \log a^{25}b^{62}.\] The answer must be expressed as $\log(b^n)$, however. We're in luck: the two different yet equal expressions for $d$ allow us to express $a$ and $b$ in terms of each other: \[\log a^2b^5 = \log a^3b^3\] \[a^2b^5 = a^3b^3\] \[a=b^2.\] Plugging in $a=b^2$, we have: \[a+11d = \log b^{50}b^{62}\] \[= \log b^{112} \Rightarrow \boxed{D}.\] ~ Jingwei325 $\smiley$

See Also

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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=2008_AMC_12A_Problems/Problem_16&oldid=178967"