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I'm new to this site and I need help on this logarithm question. I don't know how to approach this question to simplify it. $$\log_2(x^2-4)−3\log_2\frac{(x+2)}{(x-2)}>2$$

Apparently the answer is $(-\infty, -2) \cup (6, +\infty)$

Alex M.
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Mira
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2 Answers2

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First, you can use properties of logarithm. For all positive $a$, $b$, and $c\ne 1$, $$ \log_c a +\log_c b = \log_c ab. $$ Second, logarithm is defined on $(0,\infty)$. In given inequality, $x^2-4$ and $\frac{x+2}{x-2}$ must be positive.

Can you proceed from this?

choco_addicted
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We need $x^2-4>0, \dfrac{x+2}{x-2}>0$

$\implies x>2$ or $x<-2$

See Laws of Logarithms

We have $$\log_2(x-2)+\log_2(x+2)-3\{\log_2(x+2)-\log_2(x-2)\}>2$$

$$\iff\log_2(x-2)^2-\log_2(x+2)>1$$

$$\iff\log_2(x-2)^2>1+\log_2(x+2)=\log_22(x+2)$$

$$(x-2)^2>2x+4\iff x^2-6x>0\iff x(x-6)>0$$

Can you take it from here?

lab bhattacharjee
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  • and what about $x<-2$? – Mira Feb 23 '16 at 16:19
  • @mira, From the last line we need either $x>6$ or $x<0$ But ? – lab bhattacharjee Feb 23 '16 at 16:24
  • I don't dispute the final conclusion, but it seems to me you're playing fast and loose with the laws of logarithms along the way. For example, the OP's inequality makes sense when $x=-3$, but the inequality $\log_2(x-2)^2-\log_2(x+2)\gt1$ does not. – Barry Cipra Feb 23 '16 at 16:25
  • @labbhattacharjee yes, and then we combinate $x>6, x<0$ and $x<-2$ and answer $x>6, x<-2$ Thanks!!! – Mira Feb 23 '16 at 16:30