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

Revision as of 01:53, 6 January 2019

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.

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

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

@@ Line 8: / Line 8: @@
 ==Solution using [[logarithms]]==
-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>.
+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>
@@ Line 17: / Line 17: @@
 <cmath>\boxed{\frac{3}{2}}</cmath>
+==Solution using chain logarithm 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.
 == See Also ==
 {{AMC10 box|year=2005|ab=B|num-b=16|num-a=18}}
 {{MAA Notice}}

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