How to solve this inequality , when there is a irrational power?

Question

$$\frac{(x-2)(x-3)^\sqrt{2}}{(x+1)}>0$$

I'm confused with $(x-3)^\sqrt{2}$ factor.

When $x>3$ it is positive.

When $x=3$ this inequality does not hold.

When $x<3$, I am confused.

Lee Mosher · Answer 1 · 2019-09-01T14:52:01.153

9

You don't have to worry about $x < 3$ because it is not in the domain. Irrational powers are not defined when the base is negative.

You might remember that for an irrational number $a$, by definition the exponentiation function with exponent $a$ is defined by $$t^a = e^{a \ln(t)} $$ and the domain of that function is the positive axis $t > 0$.

So for your function, the domain is $x > 3$.

edited Sep 01 '19 at 14:52

answered Sep 01 '19 at 14:35

Lee Mosher

120,280

$$x=1/2$$ is not defined for instance. – Dr. Sonnhard Graubner Sep 01 '19 at 14:39
1

Thanks, I fixed the clash of variables in my answer. – Lee Mosher Sep 01 '19 at 14:53

score 5 · Answer 2 · answered Sep 01 '19 at 14:37

5

Hint: For $x<3$, $(x-3)^{\sqrt 2}$ is not defined, so you must impose the condition $x\ge 3$ from the start. Aftwerwards you may proceed as you did.

answered Sep 01 '19 at 14:37

Alexdanut

2,807

score 1 · Accepted Answer · answered Sep 01 '19 at 14:37

1

Since we have the factor$$(x-3)^{\sqrt{2}}$$ it must be $$x>3$$ then it is $$(x-3)^{\sqrt{2}}$$ a real number.

answered Sep 01 '19 at 14:37

Dr. Sonnhard Graubner

95,283

So , sir it is never a negative when $x \geq 3$ ? – Angelo Mark Sep 01 '19 at 14:42
No, it must be $x>3$ since the inequality is strict. – Dr. Sonnhard Graubner Sep 01 '19 at 14:43
$0$ is not greater than zero, so $x=3$ is not a solution. – Dr. Sonnhard Graubner Sep 01 '19 at 14:48
So numbers like $(-2)^\sqrt{3}$ are complex numbers ? – Angelo Mark Sep 01 '19 at 14:56

How to solve this inequality , when there is a irrational power?

3 Answers3