# 1997 AHSME Problems/Problem 17

## 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 $\boxed{A}$.

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