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

(→‎Solution using logarithms)
(→‎Solution using logarithm chain rule)
 
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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>?
 
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>\mathrm{(A)} 1 \qquad \mathrm{(B)} \frac{3}{2} \qquad \mathrm{(C)} 2 \qquad \mathrm{(D)} \frac{5}{2} \qquad \mathrm{(E)} 3 </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>
  
 
== Solution ==
 
== Solution ==
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<cmath> 8=7^d </cmath> <cmath>8=\left(6^c\right)^d</cmath> <cmath>8=\left(\left(5^b\right)^c\right)^d</cmath> <cmath>8=\left(\left(\left(4^a\right)^b\right)^c\right)^d</cmath> <cmath>8=4^{a\cdot b\cdot c\cdot d}</cmath> <cmath>2^3=2^{2\cdot a\cdot b\cdot c\cdot d}</cmath> <cmath>3=2\cdot a\cdot b\cdot c\cdot d</cmath> <cmath>a\cdot b\cdot c\cdot d=\boxed{\mathrm{(B)}\ \dfrac{3}{2}}</cmath>
+
<cmath>\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*}</cmath>
  
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==Solution using [[logarithms]]==
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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_56</math>, <math>c</math> as <math>\log_67</math>, and <math>d</math> as <math>\log_78</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>.  
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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>
 
  
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<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:
 +
<cmath>\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*}</cmath>
  
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<cmath>\frac{3\log2}{2\log2}</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>.
  
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<cmath>\boxed{\frac{3}{2}}</cmath>
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~solver1104
  
 
== See Also ==
 
== See Also ==
 
{{AMC10 box|year=2005|ab=B|num-b=16|num-a=18}}
 
{{AMC10 box|year=2005|ab=B|num-b=16|num-a=18}}
 +
{{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 (Problems • Answer Key • Resources)
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

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