Difference between revisions of "1997 AHSME Problems/Problem 17"
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==Problem== | ==Problem== | ||
− | A line <math> | + | A line <math>x=k</math> intersects the graph of <math>y=\log_5 x</math> and the graph of <math>y=\log_5 (x + 4)</math>. The distance between the points of intersection is <math>0.5</math>. Given that <math>k = a + \sqrt{b}</math>, where <math>a</math> and <math>b</math> are integers, what is <math>a+b</math>? |
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10 </math> | <math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10 </math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | Since the line <math>x=k</math> is vertical, we are only concerned with vertical distance. | ||
+ | |||
+ | In other words, we want to find the value of <math>k</math> for which the distance <math>|\log_5 x - \log_5 (x+4)| = \frac{1}{2}</math> | ||
+ | |||
+ | Since <math>\log_5 x</math> is a strictly increasing function, we have: | ||
+ | |||
+ | <math>\log_5 (x + 4) - \log_5 x = \frac{1}{2}</math> | ||
+ | |||
+ | <math>\log_5 (\frac{x+4}{x}) = \frac{1}{2}</math> | ||
+ | |||
+ | <math>\frac{x+4}{x} = 5^\frac{1}{2}</math> | ||
+ | |||
+ | <math>x + 4 = x\sqrt{5}</math> | ||
+ | |||
+ | <math>x\sqrt{5} - x = 4</math> | ||
+ | |||
+ | <math>x = \frac{4}{\sqrt{5} - 1}</math> | ||
+ | |||
+ | <math>x = \frac{4(\sqrt{5} + 1)}{5 - 1^2}</math> | ||
+ | |||
+ | <math>x = 1 + \sqrt{5}</math> | ||
+ | |||
+ | The desired quantity is <math>1 + 5 = 6</math>, and the answer is <math>\boxed{A}</math>. | ||
== See also == | == See also == | ||
{{AHSME box|year=1997|num-b=16|num-a=18}} | {{AHSME box|year=1997|num-b=16|num-a=18}} | ||
+ | {{MAA Notice}} |
Latest revision as of 10:31, 25 September 2016
Problem
A line intersects the graph of
and the graph of
. The distance between the points of intersection is
. Given that
, where
and
are integers, what is
?
Solution
Since the line is vertical, we are only concerned with vertical distance.
In other words, we want to find the value of for which the distance
Since is a strictly increasing function, we have:
The desired quantity is , and the answer is
.
See also
1997 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.