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

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==Solution==
 
==Solution==
  
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Since the line <math>x=k</math> is horizontal, we are only concerned with vertical distance.
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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>
 
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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== See also ==
 
== See also ==
 
{{AHSME box|year=1997|num-b=16|num-a=18}}
 
{{AHSME box|year=1997|num-b=16|num-a=18}}
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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 $\boxed{A}$.

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

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