Difference between revisions of "1997 AHSME Problems/Problem 17"

Latest revision as of 09:31, 25 September 2016

Problem

A line $x=k$ intersects the graph of $y=\log_5 x$ and the graph of $y=\log_5 (x + 4)$ . The distance between the points of intersection is $0.5$ . Given that $k = a + \sqrt{b}$ , where $a$ and $b$ are integers, what is $a+b$ ?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$

Solution

Since the line $x=k$ is vertical, we are only concerned with vertical distance.

In other words, we want to find the value of $k$ for which the distance $|\log_5 x - \log_5 (x+4)| = \frac{1}{2}$

Since $\log_5 x$ is a strictly increasing function, we have:

$\log_5 (x + 4) - \log_5 x = \frac{1}{2}$

$\log_5 (\frac{x+4}{x}) = \frac{1}{2}$

$\frac{x+4}{x} = 5^\frac{1}{2}$

$x + 4 = x\sqrt{5}$

$x\sqrt{5} - x = 4$

$x = \frac{4}{\sqrt{5} - 1}$

$x = \frac{4(\sqrt{5} + 1)}{5 - 1^2}$

$x = 1 + \sqrt{5}$

The desired quantity is $1 + 5 = 6$ , and the answer is $\boxed{A}$ .

1997 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 7: / Line 7: @@
 ==Solution==
-Since the line <math>x=k</math> is horizontal, we are only concerned with vertical distance.
+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>

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Difference between revisions of "1997 AHSME Problems/Problem 17"

Latest revision as of 09:31, 25 September 2016

Problem

Solution

See also