Difference between revisions of "2005 AMC 10B Problems/Problem 17"

(Solution using logarithms)
(Solution using chain logarithm rule)
(One intermediate revision by the same user not shown)
Line 10: Line 10:
 
We can write <math>a</math> as <math>\log_4 5</math>, <math>b</math> as <math>\log_5 6</math>, <math>c</math> as <math>\log_6 7</math>, and <math>d</math> as <math>\log_7 8</math>.  
 
We can write <math>a</math> as <math>\log_4 5</math>, <math>b</math> as <math>\log_5 6</math>, <math>c</math> as <math>\log_6 7</math>, and <math>d</math> as <math>\log_7 8</math>.  
 
We know that <math>\log_b a</math> can be rewritten as <math>\frac{\log a}{\log b}</math>, so <math>a*b*c*d=</math>
 
We know that <math>\log_b a</math> can be rewritten as <math>\frac{\log a}{\log b}</math>, so <math>a*b*c*d=</math>
<cmath>\frac{\log5}{\log4}\cdot\frac{\log6}{\log5}\cdot\frac{\log7}{\log6}\cdot\frac{\log8}{\log7}</cmath>
+
<cmath>\frac{\log5}{\log4}\cdot\frac{\log6}{\log5}\cdot\frac{\log7}{\log6}\cdot\frac{\log8}{\log7}=</cmath>
  
 
<cmath>\frac{\log8}{\log4}=</cmath>  
 
<cmath>\frac{\log8}{\log4}=</cmath>  
  
<cmath>\frac{3\log2}{2\log2}</cmath>
+
<cmath>\frac{3\log2}{2\log2}=</cmath>
  
 
<cmath>\boxed{\frac{3}{2}}</cmath>
 
<cmath>\boxed{\frac{3}{2}}</cmath>
  
==Solution using chain logarithm rule==
+
==Solution using logarithm chain rule==
 
As in solution 2, we can write <math>a</math> as <math>\log_4 5</math>, <math>b</math> as <math>\log_56</math>, <math>c</math> as <math>\log_67</math>, and <math>d</math> as <math>\log_78</math>. <math>a*b*c*d</math> is equivalent to <math>(\log_4 5)*(\log_5 6)*(\log_6 7)*(\log_7 8)</math>. Note that by the logarithm chain rule, this is equivalent to <math>\log_4 8</math>, which evaluates to <math>\frac{3}{2}</math>, so <math>\boxed{B}</math> is the answer.  
 
As in solution 2, we can write <math>a</math> as <math>\log_4 5</math>, <math>b</math> as <math>\log_56</math>, <math>c</math> as <math>\log_67</math>, and <math>d</math> as <math>\log_78</math>. <math>a*b*c*d</math> is equivalent to <math>(\log_4 5)*(\log_5 6)*(\log_6 7)*(\log_7 8)</math>. Note that by the logarithm chain rule, this is equivalent to <math>\log_4 8</math>, which evaluates to <math>\frac{3}{2}</math>, so <math>\boxed{B}</math> is the answer.  
 
~solver1104
 
~solver1104

Revision as of 22:59, 31 May 2021

Problem

Suppose that $4^a = 5$, $5^b = 6$, $6^c = 7$, and $7^d = 8$. What is $a \cdot b\cdot c \cdot d$?

$\mathrm{(A)} 1 \qquad \mathrm{(B)} \frac{3}{2} \qquad \mathrm{(C)} 2 \qquad \mathrm{(D)} \frac{5}{2} \qquad \mathrm{(E)} 3$

Solution

\[8=7^d\] \[8=\left(6^c\right)^d\] \[8=\left(\left(5^b\right)^c\right)^d\] \[8=\left(\left(\left(4^a\right)^b\right)^c\right)^d\] \[8=4^{a\cdot b\cdot c\cdot d}\] \[2^3=2^{2\cdot a\cdot b\cdot c\cdot d}\] \[3=2\cdot a\cdot b\cdot c\cdot d\] \[a\cdot b\cdot c\cdot d=\boxed{\mathrm{(B)}\ \dfrac{3}{2}}\]

Solution using logarithms

We can write $a$ as $\log_4 5$, $b$ as $\log_5 6$, $c$ as $\log_6 7$, and $d$ as $\log_7 8$. We know that $\log_b a$ can be rewritten as $\frac{\log a}{\log b}$, so $a*b*c*d=$ \[\frac{\log5}{\log4}\cdot\frac{\log6}{\log5}\cdot\frac{\log7}{\log6}\cdot\frac{\log8}{\log7}=\]

\[\frac{\log8}{\log4}=\]

\[\frac{3\log2}{2\log2}=\]

\[\boxed{\frac{3}{2}}\]

Solution using logarithm chain rule

As in solution 2, we can write $a$ as $\log_4 5$, $b$ as $\log_56$, $c$ as $\log_67$, and $d$ as $\log_78$. $a*b*c*d$ is equivalent to $(\log_4 5)*(\log_5 6)*(\log_6 7)*(\log_7 8)$. Note that by the logarithm chain rule, this is equivalent to $\log_4 8$, which evaluates to $\frac{3}{2}$, so $\boxed{B}$ is the answer. ~solver1104

See Also

2005 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png