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

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The discussion revolves around simplifying the expression [(m^3)(n^-3)]^-1 over m^-5n. Participants clarify that the original expression is not an equation and emphasize the importance of correctly applying exponent rules. The correct simplification involves distributing the negative exponent and combining like terms, ultimately leading to the expression n^3 m^{5n - 3}. Additionally, there is a brief shift to factoring a quadratic expression, with participants discussing methods like the Punnett square and the quadratic formula. The conversation highlights the common confusion between expressions and equations in algebra.
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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?
 
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I've moved this from the Tutorials Forum into the Homework Help section.

runicle said:
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?
 
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.
 
runicle said:
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.
 
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
 
So the exponent doesn't affect the base number
 
runicle said:
So the exponent doesn't affect the base number
No. There is no "base number" here anyway -- just the variables m and n.

- Warren
 
Im so confused
 
Okay wait what happens to the exponents when its (m^3)(n^2)
 
  • #10
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
 
  • #11
runicle said:
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
 
  • #12
Would it be mn^-6?
 
  • #13
Sorry i just read simplify. yes i need to simplify
 
  • #14
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
 
  • #15
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
 
  • #16
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
 
  • #17
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...
 
  • #18
runicle said:
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
 
  • #19
Well i don't know any other way
 
  • #20
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
 
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  • #21
runicle said:
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.
 
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  • #22
ksinclair13 said:
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!
 
  • #23
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
 
  • #24
VietDao29 said:
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.
 
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