# College Algebra Simplifying Exponents

1. Jun 10, 2014

### velox_xox

Hi everyone. I'm back and happy to be back. I missed math. Anyways, I am taking a college Algebra class as it has been a while, and I definitely want to make sure I have a good foundation for higher maths. But, I have a conundrum. The teacher today gave this example that I thought I had solved correctly, but I did it differently. I'm going to post how I solved it and how the teacher solved it.

My Way:
1. The problem statement, all variables and given/known data
$$-9^{-3/2}$$

2. Relevant equations
--

3. The attempt at a solution

$$-9^{-3/2}$$
$$=-(\frac{1}{9^3})^{1/2}$$
$$=-(\frac{1}{729})^{1/2}$$
$$=-(\frac{1}{27})$$ ^^
$$=±\frac{1}{27}$$

^^ Did I bend the rules here? It's a negative before that.... but then a negative of a negative ('cuz of the square root) is a positive, so plus-minus.

TEACHER'S WAY:
1. The problem statement, all variables and given/known data
$$-9^{-3/2}$$

2. Relevant equations
--

3. The attempt at a solution
$$-9^{-3/2}$$
$$=\frac{1}{-9^{3/2}}$$
$$=\frac{1}{-(9^{1/2})^3}$$
$$=\frac{-1}{(3)^3}$$
$$=\frac{-1}{27}$$
$$=±\frac{1}{27}$$

All right, so that's it. The problem solved... my way and the teacher's way. I just want to make sure the answer is correct (never hurts to verify), and that I'm not bending the rules of math again. Thanks in advance for your help.

2. Jun 10, 2014

### pasmith

By convention, $x^{1/2}$ and $\sqrt{x}$ both mean the positive root. Hence $-9^{-3/2} = -(9^{-3/2}) = -(3^{-3}) = -\frac{1}{27}$.

3. Jun 10, 2014

### BiGyElLoWhAt

I agree with pasmith, but if your teacher had the plus-minus, then the probably are using the more technical definition: $\sqrt{9} = \pm 3$

4. Jun 10, 2014

### pasmith

No, it just indicates that the teacher is unaware of the technical definition, which is $\sqrt{x} \geq 0$.

The set of numbers $x$ such that $x^2 = 9$ is $\{-3,3\}$ for which $x = \pm 3$ is an abbreviation. But $\sqrt{9} = 3$.

5. Jun 10, 2014

### BiGyElLoWhAt

What you did is fine, the only thing I want to point out is $(-9)^{-3/2}$ ,which is what your original problem looks like, is completely different from $-(9)^{-3/2}$ which is how you treated the problem.

The first solution is imaginary while the second has the real solutions that both you and your teacher arrived at.

6. Jun 10, 2014

### pasmith

By convention, $-a^b$ means $-(a^b)$. If you want $(-a)^b$ you need the brackets.

7. Jun 10, 2014

### BiGyElLoWhAt

Hmmm... I see what you're saying, (see my post about imaginary solutions) but in what I'm assuming is the context of class (please OP, correct me if I'm wrong because I very well may be)

$\sqrt{9} = x$
Solve for x as follows:
$\sqrt{9}^2=x^2$
$9 = x^2$
$0= x^2 -9$
$0 = (x+3)(x-3)$

$x= \pm 3$

8. Jun 10, 2014

### BiGyElLoWhAt

OK well I'm not going to argue over convention. I just always tend to over use brackets for clarity, soo...

Using the same logic, I could say: $x = 1$, therefore $x^2 = 1$, therefore $x^2 - 1 = 0$, therefore $(x-1)(x+1) = 0$, therefore $x=1$ or $x=-1$. Which is of course true: one of $x=1$ and $x=-1$ is true, but only one ($x=1$).