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 == | ||
| + | {{AHSME box|year=1997|num-b=16|num-a=18}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 09:31, 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 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
