Quick Math Help: Solving (m^3)(n^-3)^-1 / m^-5n - Is My Solution Correct?

[runicle]

I have the equation [(m^3)(n^-3)]^-1 over m^-5n
So far i got 1 over m^6n^6 divide by m^-5n...
Is it right or no?

 

[quantumdude]

I've moved this from the Tutorials Forum into the Homework Help section.

I have the equation [(m^3)(n^-3)]^-1 over m^-5n

That's not an equation, it's an expression.

So far i got 1 over m^6n^6 divide by m^-5n...
Is it right or no?

No, it isn't. What steps did you take to manipulate the numerator of your original expression?

 

[runicle]

First off i went to the basics I multiplied (m^3)(n^-2) and got m^6n^6
Then i inversed it to get 1 over m^6n^6, so i can divide that with m^-5. Yet, i still don't know arithmetic.

 

[quantumdude]

First off i went to the basics I multiplied (m^3)(n^-2) and got m^6n^6

That's not right. You should be using the following rule:

$$(x^ay^b)^c=x^{ac}y^{bc},$$

which should be in your textbook.

 

[chroot]

What are you trying to do? Simplify the expression?

Note that $m^3 \cdot n^{-3}$ does not equal $m^6 n^6$.

Where did the extra powers of m come from? Where did the negative powers of n go?

- Warren

 

[runicle]

So the exponent doesn't affect the base number

 

[chroot]

So the exponent doesn't affect the base number

No. There is no "base number" here anyway -- just the variables m and n.

- Warren

 

[runicle]

Im so confused

 

[runicle]

Okay wait what happens to the exponents when its (m^3)(n^2)

 

[chroot]

Did you see what Hurkyl posted? The general rule of "exponent distribution:"

$$(x^ay^b)^c=x^{ac}y^{bc}$$

You have an example of this here:

$$(m^3 n^{-3})^{-1} = m^{3 \cdot -1}n^{-3 \cdot -1} = m^{-3} n^{3}$$

Again, I'll ask you: what are you trying to do? Simplify the expression?

- Warren

 

[chroot]

Okay wait what happens to the exponents when its (m^3)(n^2)

Nothing at all "happens" to the exponents -- the bases are different, and thus they are completely unrelated to each other.

- Warren

 

[runicle]

Would it be mn^-6?

 

[runicle]

Sorry i just read simplify. yes i need to simplify

 

[chroot]

No. Now you're getting me a bit confused. First you told us the problem was this:

(m^3)(n^-3)]^-1 over m^-5n

and now you're talking about (m^3)(n^2). Which is it?

Please repeat the problem, taking care to type it exactly as shown in your homework. Also, please tell us exactly what you're trying to do: are you just supposed to simplify the expression?

- Warren

 

[chroot]

No. Now you're getting me a bit confused. First you told us the problem was this:

(m^3)(n^-3)]^-1 over m^-5n

and now you're talking about (m^3)(n^2). Which is it?

Please repeat the problem, taking care to type it exactly as shown in your homework. Also, please tell us exactly what you're trying to do: are you just supposed to simplify the expression?

- Warren

 

[runicle]

Nevermind i got it I'm simplifying it It's n^2m^5 over m^3n

The question was simplify [(m^3)(n^-3)]^-1 over m^-5n

 

[runicle]

I got another question I have to factor 3x^2 -13x-10
So far i get this far
3x^2 -13x-10
3x^2+2x-15x-10
x(3x+2)-5(3x+2)
What do i do next...

 

[chroot]

Nevermind i got it I'm simplifying it It's n^2m^5 over m^3n

The question was simplify [(m^3)(n^-3)]^-1 over m^-5n

I'm afraid that

$\frac{(m^3 n^{-3})^{-1}}{m^{-5n}}$

does not simplify to

$\frac{n^2 m^5}{m^{3n}}$

- Warren

 

[runicle]

Well i don't know any other way

 

[chroot]

Let's do this in steps.

First, simplify the numerator of

$\frac{(m^3 n^{-3})^{-1}}{m^{-5n}}$

The -1 exponent "distributes", multiplying the exponents on m and n:

$(m^3 n^{-3})^{-1} = m^{-3}n^3}$

So the entire expression becomes:

$\frac{m^{-3}n^3}{m^{-5n}}$

Now, factors with negative exponents can be "flipped" across the division line, as can be seen by multiplying both sides by the same factor with the exponent positive rather than negative:

$\frac{m^{-3}n^3}{m^{-5n}} = m^{-3}n^3 m^{5n}$

One negative-exponent factor went up, while the other negative-exponent factor I left alone.

Finally, multiplyiing two factors (with the same base) is the same as adding their exponents:

$n^3 m^{5n - 3}$

If you do not understand any step of this, please let me know.

- Warren

 

[ksinclair13]

I got another question I have to factor 3x^2 -13x-10
So far i get this far
3x^2 -13x-10
3x^2+2x-15x-10
x(3x+2)-5(3x+2)
What do i do next...

Your final answer will be:

0 = (x - 5)(3x + 2)

You should see that multiplying (3x + 2) by x and then adding that to -5(3x + 2) is the same as just multiplying by (x - 5).

I find it easiest to make a punett square if I can't figure it out in my head. If you can't figure it out using a punett square, then try using the quadratic formula; I try and stay away from that though because it can be messy.

 

[VietDao29]

Your final answer will be:

0 = (x - 5)(3x + 2)

No it isn't. It's an expression that need to be factored, not an equation!

 

[runicle]

I know what I'm doing... I always mix up the terms equation and expression. I know now 2^-3 is the same thing as 1 over 2^3 so now you can understand how i did it

 

[ksinclair13]

No it isn't. It's an expression that need to be factored, not an equation!

Whoa! I must have read that too quickly. Sorry if I misled you runicle.