# How to solve this logarithmic inequality

• December 28th 2011, 06:43 AM
LiberChronicon
How to solve this logarithmic inequality
Hi all,
How can I solve this?
http://www4a.wolframalpha.com/Calcul...=62&w=233&h=30

I tried with log_a (b) - log_a (c) = log_a (b/c) but the solution is (13+5x)/(x^2+5x+4)>=0 ...how is this possible?

• December 28th 2011, 06:58 AM
Plato
Re: How to solve this logarithmic inequality
Quote:

Originally Posted by LiberChronicon
Hi all,
How can I solve this?
http://www4a.wolframalpha.com/Calcul...=62&w=233&h=30
I tried with log_a (b) - log_a (c) = log_a (b/c) but the solution is (13+5x)/(x^2+5x+4)>=0 ...how is this possible?

You want to solve $\frac{x^2-9}{x^2+5x+4)}\le 1$.
• December 28th 2011, 07:00 AM
HallsofIvy
Re: How to solve this logarithmic inequality
Okay, as you say, $log_{1/3}(x^2- 9)- log_{1/3}(x^2+ 5x+ 4)\ge 0$ is the same as $log_{1/3}\left(\frac{x^2- 9}{x^2+ 5x+ 4}\right)\ge 0$.

Further $log_a(x)\ge 0$, for any a< 1, if and only if $x\le 1$. That is, your inequality is the same as $\frac{x^2- 9}{x^2+ 5x+ 4}\le 1$.

Subtracting 1 from both sides, $\frac{x^2- 9}{x^2+ 5x+ 4}- 1= \frac{x^2- 9- (x^2+ 5x+ 4)}{x^2+ 5x+ 4}= \frac{-5x- 13}{x^2+ 5x+ 5}\le 0$.

Now, multiply both sides by -1.