Maths Olympiad Prep

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Algebra Difficulty 5.2 AIME, harder Find the answer

Solve the following equation:

log⁡[1x+96x2(x2−1)]=log⁡(x+1)+log⁡(x+2)+log⁡(x+3)−2log⁡x−log⁡(x2−1) \log \left[\frac{1}{x}+\frac{96}{x^{2}\left(x^{2}-1\right)}\right]=\log (x+1)+\log (x+2)+\log (x+3)-2 \log x-\log \left(x^{2}-1\right)

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Our equation can also be written as:

1x+96x2(x2−1)=(x+1)(x+2)(x+3)x2(x2−1) \frac{1}{x}+\frac{96}{x^{2}\left(x^{2}-1\right)}=\frac{(x+1)(x+2)(x+3)}{x^{2}\left(x^{2}-1\right)}

or, after rearranging:

x2+2x=15 x^{2}+2 x=15

from which

x1=3,x2=−5 x_{1}=3, x_{2}=-5

(Julius Weber, Beszterczebánya.)

The problem was also solved by: Z. Harsányi, P. Heimlich, Gy. Jánosy, J. Kiss, E. Makó, Sarolta Petrik, S. Pichler, E. Sárközy, Gy. Schlesinger, Gy. Schwarz, Gy. Schuster, I. Szécsi, D. Szőke, B. Tóth.

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