Exponential Inequalities

When we solve inequalities of the form `a^(f(x))>a^(g(x))` we need to remember, that exponential function `y=a^x` is increasing when a>1 and decreasing when 0<a<1.

Therefore, when a>1, `a^(f(x))>a^(g(x))` is equivalent to the inequality `f(x)>g(x)`. If 0<a<1, then `a^(f(x))>a^(g(x))` is euqivalent to `f(x)<g(x)`.

Example 1. Solve `2^(3x+7)<2^(2x-1)`.

Here, base of exponent is greater than 1 (it is 2), therefore, equivalent inequality is `3x+7<2x-1`. Solving it, we obtain that `x<-8`.

Example 2. Solve `(0.04)^(5x-x^2-8)<=625`.

Since `625=25^2=((1/(25))^-1)^2=(1/(25))^(-2)=(0.04)^(-2)`, then we can rewrite inequality as `(0.04)^(5x-x^2-8)<=(0.04)^-2`.

Since `0<0.04<1`, then equivalent inequality is `5x-x^2-8>=-2`. From this we obtain that

`-x^2+5x-6>=0`, `x^2-5x+6<=0`, `(x-2)(x-3)<=0`.

Using method of intervals we obtain that solution is `2<=x<=3`.