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

 
(Solution using logarithm chain rule)
 
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== Problem ==
 
== Problem ==
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Suppose that <math>4^a = 5</math>, <math>5^b = 6</math>, <math>6^c = 7</math>, and <math>7^d = 8</math>. What is <math>a \cdot b\cdot c \cdot d</math>?
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<math>\textbf{(A) } 1 \qquad \textbf{(B) } \frac{3}{2} \qquad \textbf{(C) } 2 \qquad \textbf{(D) } \frac{5}{2} \qquad \textbf{(E) } 3 </math>
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== Solution ==
 
== Solution ==
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<cmath>\begin{align*}
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8&=7^d \
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8&=\left(6^c\right)^d\
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8&=\left(\left(5^b\right)^c\right)^d\
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8&=\left(\left(\left(4^a\right)^b\right)^c\right)^d\
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8&=4^{a\cdot b\cdot c\cdot d}\
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2^3&=2^{2\cdot a\cdot b\cdot c\cdot d}\
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3&=2\cdot a\cdot b\cdot c\cdot d\
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a\cdot b\cdot c\cdot d&=\boxed{\textbf{(B) }\dfrac{3}{2}}\
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\end{align*}</cmath>
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==Solution 2 ([[logarithms]])==
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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>.
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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*}
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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}} \
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a\cdot b \cdot c \cdot d &= \frac{\log8}{\log4} \
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a\cdot b \cdot c \cdot d &= \frac{3\cancel{\log2}}{2\cancel{\log2}} \
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a\cdot b \cdot c \cdot d &= \boxed{\textbf{(B) }\frac{3}{2}} \
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\end{align*}</cmath>
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==Solution 3 (logarithm chain rule)==
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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>.
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~solver1104
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== See Also ==
 
== See Also ==
*[[2005 AMC 10B Problems]]
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{{AMC10 box|year=2005|ab=B|num-b=16|num-a=18}}
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{{MAA Notice}}

Latest revision as of 14:30, 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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