2005 AMC 10B Problems/Problem 17: Difference between revisions

Revision as of 21:58, 31 May 2021

Problem

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

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

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

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

@@ Line 12: / Line 12: @@
 <cmath>\frac{\log5}{\log4}\cdot\frac{\log6}{\log5}\cdot\frac{\log7}{\log6}\cdot\frac{\log8}{\log7}</cmath>
-<cmath>\frac{\log8}{\log4}</cmath>
+<cmath>\frac{\log8}{\log4}=</cmath>
 <cmath>\frac{3\log2}{2\log2}</cmath>

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