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

Revision as of 12:56, 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 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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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