# Logarithmic Inequalities

When we solve inequalities of the form $log_a(f(x))>log_a(g(x))$ we need to remember, that logarithmic function $y=log_a(x)$ is increasing when a>1 and decreasing when 0<a<1. Also, we need to remember, that logarithmic function is defined

Therefore, we have 2 cases:

1. If a>1, then $log_a(f(x))>log_a(g(x))$ is equivalent to ${(f(x)>g(x)),(f(x)>0),(g(x)>0):}$ which is equivalent to ${(f(x)>g(x)),(g(x)>0):}$, because if g(x)>0 and f(x)>g(x), then f(x)>0 automatically.
2. If 0<a<1, then $log_a(f(x))>log_a(g(x))$ is equivalent to ${(f(x)<g(x)),(f(x)>0),(g(x)>0):}$ which is equivalent to ${(f(x)<g(x)),(f(x)>0):}$, because if f(x)>0 and f(x)<g(x), then g(x)>0 automatically.

Example 1. Solve $log_(0.5)(2x+59)> -2$.

Since $-2=log_(0.5)(4)$, then we can rewrite inequality as $log_(0.5)(2x+59)>log_(0.5)(4)$.

Now, since $0<0.5<1$, then inequality is equivalent to the system ${(2x+59<4),(2x+59>0):}$.

This system is equivalent to the system ${(x<-55/2),(x> -59/2):}$.

Therefore, solution is $-59/2<x< -55/2$.

Example 2. Solve $text(lg)(x+2)<2-text(lg)(2x-6)$. (recall that $text(lg)$ is $log_(10)$).

Note, that arguments of logarithms should be positive.

Now, we need to transform inequality:

$text(lg)(x+2)+text(lg)(2x-6)<2$;

$text(lg)((x+2)(2x-6))<2$;

$text(lg)((x+2)(2x-6))<text(lg)(100)$.

Since base of logarithm 10>1, then equivalent system is ${((x+2)(2x-6)<100),(x+2>0),(2x-6>0):}$.

This system can be rewritten as ${(x^2-x-56<0),(x> -2),(x>3):}$.

Equivalent of this system is system ${((x+7)(x-8)<0),(x>3):}$, i.e. ${(-7<x<8),(x>3):}$

Using coordinate line (see figure) we obtain that solution of this system and, thus, initial inequality is interval (3,8).