Difference between revisions of "2005 AMC 10B Problems/Problem 17"
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== 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 \cdot b\cdot c \cdot d</math>? | ||
| + | |||
| + | <math>\textbf{(A) } 1 \qquad \textbf{(B) } \frac{3}{2} \qquad \textbf{(C) } 2 \qquad \textbf{(D) } \frac{5}{2} \qquad \textbf{(E) } 3 </math> | ||
| + | |||
== Solution == | == Solution == | ||
| + | <cmath>\begin{align*} | ||
| + | 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{\textbf{(B) }\dfrac{3}{2}}\\ | ||
| + | \end{align*}</cmath> | ||
| + | |||
| + | ==Solution 2 ([[logarithms]])== | ||
| + | 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 we have: | ||
| + | <cmath>\begin{align*} | ||
| + | a\cdot b \cdot c \cdot d &= \frac{\cancel{\log5}}{\log4}\cdot\frac{\cancel{\log6}}{\cancel{\log5}}\cdot\frac{\cancel{\log7}}{\cancel{\log6}}\cdot\frac{\log8}{\cancel{\log7}} \\ | ||
| + | a\cdot b \cdot c \cdot d &= \frac{\log8}{\log4} \\ | ||
| + | a\cdot b \cdot c \cdot d &= \frac{3\cancel{\log2}}{2\cancel{\log2}} \\ | ||
| + | a\cdot b \cdot c \cdot d &= \boxed{\textbf{(B) }\frac{3}{2}} \\ | ||
| + | \end{align*}</cmath> | ||
| + | |||
| + | ==Solution 3 (logarithm chain 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\cdot b\cdot c\cdot d</math> is equivalent to <math>(\log_4 5)\cdot (\log_5 6)\cdot (\log_6 7)\cdot (\log_7 8)</math>. By the logarithm chain rule, this is equivalent to <math>\log_4 8</math>, which evaluates to <math>\boxed{\textbf{(B) }\frac{3}{2}}</math>. | ||
| + | |||
| + | ~solver1104 | ||
| + | |||
== See Also == | == See Also == | ||
| − | + | {{AMC10 box|year=2005|ab=B|num-b=16|num-a=18}} | |
| + | {{MAA Notice}} | ||
Latest revision as of 14:30, 16 December 2021
Contents
Problem
Suppose that 
, 
, 
, and 
. What is 
?
Solution
Solution 2 (logarithms)
We can write 
as 
, 
as 
, 
as 
, and 
as 
.
We know that 
can be rewritten as 
, so we have:
Solution 3 (logarithm chain rule)
As in Solution 2, we can write as , as 
, as 
, and as 
. 
is equivalent to 
. By the logarithm chain rule, this is equivalent to 
, which evaluates to 
.
~solver1104
See Also
| 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. 
