Exponential inequation

[shervlad]

\(\displaystyle \eqalign{
& {\left( {{1 \over 3}} \right)^{\sqrt {x + 2} }} < {3^{ - x}} \cr
& DVA:\,\,x + 2 \ge 0 \Leftrightarrow x \ge - 2 \cr
& {\left( {{1 \over 3}} \right)^{\sqrt {x + 2} }} < {\left( {{1 \over 3}} \right)^x} \cr
& \sqrt {x + 2} > x \cr} \)
here I am stuck, if I square the both sides, it gives me x e (-1;2) but the solutions (at the end of the book) is [-2;1), how do I get to that solution?

 

[Deleted member 4993]

\(\displaystyle \eqalign{
& {\left( {{1 \over 3}} \right)^{\sqrt {x + 2} }} < {3^{ - x}} \cr
& DVA:\,\,x + 2 \ge 0 \Leftrightarrow x \ge - 2 \cr
& {\left( {{1 \over 3}} \right)^{\sqrt {x + 2} }} < {\left( {{1 \over 3}} \right)^x} \cr
& \sqrt {x + 2} > x \cr} \)
here I am stuck, if I square the both sides, it gives me x e (-1;2) but the solutions (at the end of the book) is [-2;1), how do I get to that solution?

Square both sides:

x+2 > x²

x² - x - 2 < 0

(x + 1)(x-2) < 0

Now think and continue....

It might help you, if you plot y = (x + 1)(x-2) and stare at it a bit

 

[shervlad]

Square both sides:

x+2 > x²

x² - x - 2 < 0

(x + 1)(x-2) < 0

Now think and continue....

It might help you, if you plot y = (x + 1)(x-2) and stare at it a bit

\(\displaystyle \eqalign{
& {\left( {{1 \over 3}} \right)^{\sqrt {x + 2} }} < {3^{ - x}} \cr
& \sqrt {x + 2} > x \cr
& for\, - 2 \le x \le 0 \cr
& \sqrt {x + 2} > x \cr
& x \ge 0,\,\,x + 2 > {x^2} \cr
& {x^2} - x - 2 < 0 \cr
& x \in \left( { - 1;2} \right) \cr
& x \in \left[ { - 2;2} \right) \cr}
\)
is this right?

 

[HallsofIvy]

\(\displaystyle \eqalign{
& {\left( {{1 \over 3}} \right)^{\sqrt {x + 2} }} < {3^{ - x}} \cr
& \sqrt {x + 2} > x \cr
& for\, - 2 \le x \le 0 \cr
& \sqrt {x + 2} > x \cr
& x \ge 0,\,\,x + 2 > {x^2} \cr
& {x^2} - x - 2 < 0 \cr
& x \in \left( { - 1;2} \right) \cr
& x \in \left[ { - 2;2} \right) \cr}
\)
is this right?

They can't possibly both be right!
You were told before to look at \(\displaystyle x^2- x- 2= (x- 2)(x+ 1)< 0\)
A product of two numbers is negative if and only if one factor is negative and the other is positive. That is, either x- 2> 0 and x+ 1< 0 or x- 2< 0 and x+ 1> 0.
Of course, x- 2> 0 is the same as x> 2 an then x+1 cannot be negative. if x- 2< 0, then x< 2 and x+ 1> 0 if and only if x> -1. We must have -1< x< 2.

I don't know where you got that "-2" from.

 

[Deleted member 4993]

\(\displaystyle \eqalign{
& {\left( {{1 \over 3}} \right)^{\sqrt {x + 2} }} < {3^{ - x}} \cr
& \sqrt {x + 2} > x \cr
& for\, - 2 \le x \le 0 \cr
& \sqrt {x + 2} > x \cr
& x \ge 0,\,\,x + 2 > {x^2} \cr
& {x^2} - x - 2 < 0 \cr
& x \in \left( { - 1;2} \right) \cr
& x \in \left[ { - 2;2} \right) \cr}
\)
is this right?

If you had plotted the function y = (x+1)(x-2) → you would have seen that your answer is incorrect.

 

[shervlad]

If you had plotted the function y = (x+1)(x-2) → you would have seen that your answer is incorrect.

yes, but try and put -2 in the original equation and you will see that it IS a solution

 

[daon2]

yes, but try and put -2 in the original equation and you will see that it IS a solution

Yes, it is. You have two solution sets \(\displaystyle A\) and \(\displaystyle B\) with \(\displaystyle A\subseteq B\). Hence the answer is the larger set, \(\displaystyle B\)

 

[daon2]

I think I see where the confusion in this thread is coming from. It is not true that

\(\displaystyle a > b \implies a^2 > b^2\)

The error in calculation is coming from where "b", in this case the function \(\displaystyle y=x\), takes on negative values of absolute value greater than "a", the function \(\displaystyle y=\sqrt{x+2}\).