Difference between revisions of "2005 AMC 10B Problems/Problem 17"
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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>? | 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>\ | + | <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 == |
Revision as of 12:53, 16 December 2021
Contents
Problem
Suppose that ,
,
, and
. What is
?
Solution
Solution using logarithms
We can write as
,
as
,
as
, and
as
.
We know that
can be rewritten as
, so
Solution using logarithm chain rule
As in solution 2, we can write as
,
as
,
as
, and
as
.
is equivalent to
. Note that by the logarithm chain rule, this is equivalent to
, which evaluates to
, so
is the answer.
~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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.