# Hi all, I am very new to Mathematics and I am studying inequalities currently... I have problem in solving exponential inequalities. I mean I can solve the very simple one with help of logarithm...

Hi all,

I am very new to Mathematics and I am studying inequalities currently... I have problem in solving exponential inequalities.

I mean I can solve the very simple one with help of logarithm but not more complex ones... for example:

1) (2/3)^(x^2-4x) =< 1

&

2) 2^(x^2-16) >= 1

I really appreciate if someone tell me how to solve these kinds of exponential inequalities step by step.. and please consider that you are helping a dummy!!! Advance explanation is really appreciated! Thankssss...

embizze | Certified Educator

In both of these cases it helps to consider an easier problem:

(1) Solve `2^(x^2-16)>=1`

Consider `2^k >=1` . In this case it is clear that `k>=0` is a solution. (Take `log_2` of both sides)

Thus `x^2-16>=0` will yield the solution. In order to solve this you realize that `x^2-16` is a quadratic whose graph is a parabola, with vertex (0,0) and opening up. The zeros can be found by factoring: (x+4)(x-4)=0 so x=-4 or 4. The quadratic is greater than 0 on `(-oo,-4] uu [4,oo)` .

Thus the solution is `x<=-4` or `x>=4` . The graph of `2^(x^2-16):`

(2) Solve `(2/3)^(x^2-4x)<=1`

Consider `(2/3)^k<=1` . It is clear that `k>=0` provides the solutions.(He``re the base is less than one)

Thus `x^2-4x>=0` yields the solutions. `x(x-4)>=0` : The zeros of the quadratic are at 0 and 4, and the function is positive on `(-oo,0]uu[4,oo)`

The solution is`` `x<=0` or `x>=4` .

The graph of `(2/3)^(x^2-4x):`