Maths Olympiad Prep

Track / Stage 3 / 125 of 260 #125 of 1964

Problem 125

AMC 10/12, early questions
Algebra Difficulty 3.3 Multiple choice

A line x=kx=k intersects the graph of y=log5xy=\log_5 x and the graph of y=log5(x+4)y=\log_5 (x + 4). The distance between the points of intersection is 0.50.5. Given that k=a+bk = a + \sqrt{b}, where aa and bb are integers, what is a+ba+b?

Pick one

Official solution

Since the line x=kx=k is vertical, we are only concerned with vertical distance.
In other words, we want to find the value of kk for which the distance log5xlog5(x+4)=12|\log_5 x - \log_5 (x+4)| = \frac{1}{2}
Since log5x\log_5 x is a strictly increasing function, we have:
log5(x+4)log5x=12\log_5 (x + 4) - \log_5 x = \frac{1}{2}
log5(x+4x)=12\log_5 (\frac{x+4}{x}) = \frac{1}{2}
x+4x=512\frac{x+4}{x} = 5^\frac{1}{2}
x+4=x5x + 4 = x\sqrt{5}
x5x=4x\sqrt{5} - x = 4
x=451x = \frac{4}{\sqrt{5} - 1}
x=4(5+1)512x = \frac{4(\sqrt{5} + 1)}{5 - 1^2}
x=1+5x = 1 + \sqrt{5}
The desired quantity is 1+5=61 + 5 = 6, and the answer is A\boxed{A}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.