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

(Solution)
(Solution using logarithm chain rule)
 
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2^3&=2^{2\cdot 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\\
 
3&=2\cdot a\cdot b\cdot c\cdot d\\
a\cdot b\cdot c\cdot d&=\boxed{\textbf{(B) }\ \dfrac{3}{2}}\\
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a\cdot b\cdot c\cdot d&=\boxed{\textbf{(B) }\dfrac{3}{2}}\\
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  
==Solution using [[logarithms]]==
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==Solution 2 ([[logarithms]])==
 
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>
 
<cmath>\frac{\log5}{\log4}\cdot\frac{\log6}{\log5}\cdot\frac{\log7}{\log6}\cdot\frac{\log8}{\log7}=</cmath>
 
  
<cmath>\frac{\log8}{\log4}=</cmath>  
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We know that <math>\log_b a</math> can be rewritten as <math>\frac{\log a}{\log b}</math>, so we have:
 
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<cmath>\begin{align*}
<cmath>\frac{3\log2}{2\log2}=</cmath>
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a\cdot b \cdot c \cdot d &= \frac{\cancel{\log5}}{\log4}\cdot\frac{\cancel{\log6}}{\cancel{\log5}}\cdot\frac{\cancel{\log7}}{\cancel{\log6}}\cdot\frac{\log8}{\cancel{\log7}} \\
 +
a\cdot b \cdot c \cdot d &= \frac{\log8}{\log4} \\
 +
a\cdot b \cdot c \cdot d &= \frac{3\cancel{\log2}}{2\cancel{\log2}} \\
 +
a\cdot b \cdot c \cdot d &= \boxed{\textbf{(B) }\frac{3}{2}} \\
 +
\end{align*}</cmath>
  
<cmath>\boxed{\frac{3}{2}}</cmath>
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==Solution 3 (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\cdot b\cdot c\cdot d</math> is equivalent to <math>(\log_4 5)\cdot (\log_5 6)\cdot (\log_6 7)\cdot (\log_7 8)</math>. By the logarithm chain rule, this is equivalent to <math>\log_4 8</math>, which evaluates to <math>\boxed{\textbf{(B) }\frac{3}{2}}</math>.
  
==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.
 
 
~solver1104
 
~solver1104
  

Latest revision as of 15:30, 16 December 2021

Problem

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

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

Solution

\begin{align*} 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{\textbf{(B) }\dfrac{3}{2}}\\ \end{align*}

Solution 2 (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 we have: \begin{align*} a\cdot b \cdot c \cdot d &= \frac{\cancel{\log5}}{\log4}\cdot\frac{\cancel{\log6}}{\cancel{\log5}}\cdot\frac{\cancel{\log7}}{\cancel{\log6}}\cdot\frac{\log8}{\cancel{\log7}} \\ a\cdot b \cdot c \cdot d &= \frac{\log8}{\log4} \\ a\cdot b \cdot c \cdot d &= \frac{3\cancel{\log2}}{2\cancel{\log2}} \\ a\cdot b \cdot c \cdot d &= \boxed{\textbf{(B) }\frac{3}{2}} \\ \end{align*}

Solution 3 (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 $\log_78$. $a\cdot b\cdot c\cdot d$ is equivalent to $(\log_4 5)\cdot (\log_5 6)\cdot (\log_6 7)\cdot (\log_7 8)$. By the logarithm chain rule, this is equivalent to $\log_4 8$, which evaluates to $\boxed{\textbf{(B) }\frac{3}{2}}$.

~solver1104

See Also

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

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