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

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<cmath> 8=7^d </cmath> <cmath>8=\left(6^c\right)^d</cmath> <cmath>8=\left(\left(5^b\right)^c\right)^d</cmath> <cmath>8=\left(\left(\left(4^a\right)^b\right)^c\right)^d</cmath> <cmath>8=4^{a\cdot b\cdot c\cdot d}</cmath> <cmath>2^3=2^{2\cdot a\cdot b\cdot c\cdot d}</cmath> <cmath>3=2\cdot a\cdot b\cdot c\cdot d</cmath> <cmath>a\cdot b\cdot c\cdot d=\boxed{\mathrm{(B)}\ \dfrac{3}{2}}</cmath>
 
<cmath> 8=7^d </cmath> <cmath>8=\left(6^c\right)^d</cmath> <cmath>8=\left(\left(5^b\right)^c\right)^d</cmath> <cmath>8=\left(\left(\left(4^a\right)^b\right)^c\right)^d</cmath> <cmath>8=4^{a\cdot b\cdot c\cdot d}</cmath> <cmath>2^3=2^{2\cdot a\cdot b\cdot c\cdot d}</cmath> <cmath>3=2\cdot a\cdot b\cdot c\cdot d</cmath> <cmath>a\cdot b\cdot c\cdot d=\boxed{\mathrm{(B)}\ \dfrac{3}{2}}</cmath>
  
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==Solution using [[logarithms]]==
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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>.
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We know that <math>log_b a</math> can be rewritten as <math>\frac{\log a}{\log b}, so </math>a*b*c*d=$
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<cmath>\frac{\log5}{\log4}\cdot\frac{\log6}{\log5}\cdot\frac{\log7}{\log6}\cdot\frac{\log8}{\log7}</cmath>
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<cmath>\frac{\log8}{\log4}</cmath>
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<cmath>\frac{3\log2}{2\log2}</cmath>
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<cmath>\boxed{\frac{3}{2}}</cmath>
  
 
== See Also ==
 
== See Also ==
 
{{AMC10 box|year=2005|ab=B|num-b=16|num-a=18}}
 
{{AMC10 box|year=2005|ab=B|num-b=16|num-a=18}}

Revision as of 22:30, 24 May 2013

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 $\log_56$, $c$ as $\log_67$, and $d$ as $\log_78$. We know that $log_b a$ can be rewritten as $\frac{\log a}{\log b}, so$a*b*c*d=$ \[\frac{\log5}{\log4}\cdot\frac{\log6}{\log5}\cdot\frac{\log7}{\log6}\cdot\frac{\log8}{\log7}\]

\[\frac{\log8}{\log4}\]

\[\frac{3\log2}{2\log2}\]

\[\boxed{\frac{3}{2}}\]

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