# 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).