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).logarithmic inequalities