Answer to Question #217185 in Algebra for Gartiffer

Question

The real solutions of the inequality log₂(x²-1)-log₂(x-2)>2 are?

Answer 1

It follows that $x^2-1>0, x-2>0$ and $\log_2(\frac{x^2-1}{x-2})>\log_24.$ Then $x^2>1, x>2$ and $\frac{x^2-1}{x-2}>4.$ The last inequality is equivalent to $\frac{x^2-1}{x-2}-4>0,$ which is equivalent to $\frac{x^2-4x+7}{x-2}>0,$ and hence to $\frac{(x-2)^2+3}{x-2}>0.$ Since $(x-2)^2+3>0$ for any real x$,$ we conclude that the last inequality is equivalent to the inequality $x>2.$ Taking into account that the inequality $x^2>1$ is equivalent to $|x|>1,$ and hence $x\in(-\infty,-1)\cup(1,+\infty),$ we conclude that the real solutions of the inequality $\log_2(x^2-1)-\log_2(x-2)>2$ are all real numbers that are greater than 2.