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 (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 10 Problems and Solutions

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