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

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

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AMC 10 Problems and Solutions

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@@ Line 16: / Line 16: @@
 \end{align*}</cmath>
-==Solution using [[logarithms]]==
+==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 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>
+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>
-<cmath>\frac{3\log2}{2\log2}=</cmath>
+==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>.
-<cmath>\boxed{\frac{3}{2}}</cmath>
-==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

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