# 2005 AMC 10B Problems/Problem 17

## Problem

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

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

## Solution

$$8=7^d$$ $$8=\left(6^c\right)^d$$ $$8=\left(\left(5^b\right)^c\right)^d$$ $$8=\left(\left(\left(4^a\right)^b\right)^c\right)^d$$ $$8=4^{a\cdot b\cdot c\cdot d}$$ $$2^3=2^{2\cdot a\cdot b\cdot c\cdot d}$$ $$3=2\cdot a\cdot b\cdot c\cdot d$$ $$a\cdot b\cdot c\cdot d=\boxed{\mathrm{(B)}\ \dfrac{3}{2}}$$

## 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 chain logarithm 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

## See Also

 2005 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 16 Followed byProblem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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