College Algebra Simplifying Exponents

[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:

Homework Statement

$$-9^{-3/2}$$

Homework Equations

--

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:

Homework Statement

$$-9^{-3/2}$$

Homework Equations

--

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.

 

[pasmith]

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:

Homework Statement

$$-9^{-3/2}$$

Homework Equations

--

The Attempt at a Solution

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

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

 

[BiGyElLoWhAt]

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

 

[pasmith]

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

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 $\sqrt{9} = 3$.

 

[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.

 

[pasmith]

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.

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

 

[BiGyElLoWhAt]

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 $\sqrt{9} = 3$.

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:
$\sqrt{9}^2=x^2$
$9 = x^2$
$0= x^2 -9$
$0 = (x+3)(x-3)$
∴
$x= \pm 3$

 

[BiGyElLoWhAt]

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

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

I bid you adieu

 

[jbunniii]

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:
$\sqrt{9}^2=x^2$

The first line implies the second, but not vice versa.

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 ($x=1$).