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

 
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
Suppose that <math>4^a = 5</math>, <math>5^b = 6</math>, <math>6^c = 7</math>, and <math>7^d = 8</math>. What is <math>a * b * c * d</math>?
 +
 +
<math>\mathrm{(A)} 1 \qquad \mathrm{(B)} \frac{3}{2} \qquad \mathrm{(C)} 2 \qquad \mathrm{(D)} \frac{5}{2} \qquad \mathrm{(E)} 3 </math>
 +
 
== Solution ==
 
== Solution ==
 +
<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>
 +
 +
 
== See Also ==
 
== See Also ==
*[[2005 AMC 10B Problems]]
+
{{AMC10 box|year=2005|ab=B|num-b=16|num-a=18}}

Revision as of 13:45, 6 July 2011

Problem

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

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

Solution

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


See Also

2005 AMC 10B (ProblemsAnswer KeyResources)
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