Maths Olympiad Prep

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Problem 1178

AMC 12 late, AIME early
Algebra Difficulty 4.9 Prove it PMO Area Stage · Philippines

The points (3,m)(3, m), (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are on the graph of the function f(x)=log⁡axf(x) = \log_{a} x. If y1+y2=2my_1 + y_2 = 2m, find the value of x1x2x_1 x_2.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

Solution:

Since (3,m)(3, m) is on the graph of f(x)=log⁡axf(x) = \log_{a} x, we have:
m=log⁡a3 m = \log_{a} 3

Similarly, (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are on the graph, so:
y1=log⁡ax1y2=log⁡ax2 y_1 = \log_{a} x_1 \\ y_2 = \log_{a} x_2
Given y1+y2=2my_1 + y_2 = 2m, so:
log⁡ax1+log⁡ax2=2log⁡a3 \log_{a} x_1 + \log_{a} x_2 = 2 \log_{a} 3
Using the property log⁡ax1+log⁡ax2=log⁡a(x1x2)\log_{a} x_1 + \log_{a} x_2 = \log_{a} (x_1 x_2):
log⁡a(x1x2)=log⁡a32=log⁡a9 \log_{a} (x_1 x_2) = \log_{a} 3^2 = \log_{a} 9
Therefore,
x1x2=9 x_1 x_2 = 9

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