# Simplify the following problem with exponents

1. Dec 8, 2008

### TayTayDatDude

1. The problem statement, all variables and given/known data
Simplify. Express final answers with positive exponents.

((64m^-36)(n^-15)(p^9)) ^ (2/3)

2. Relevant equations
$$\sqrt[3]{}$$ ((64m^-36)(n^-15)(p^9)) ^ 2

3. The attempt at a solution

((4m^-36)(n^-15)(p^9))^2
= (16m^-36)(n^-15)(p^9)

= 1/(16m^36)(n^15)(p^-9)

I can't seem to rid the negative exponents. It would be amazing if someone could find me a page which shows the laws and helps me with this kinda stuff. thanks.

2. Dec 8, 2008

### Mentallic

Re: Exponents

ok you have the wrong idea of how the power outside the entire expression works.

$$(ab)^n=a^nb^n$$

So: $$[(64m^{-36})(n^{-15})(p^9)]^{\frac{2}{3}}$$
becomes
$$(64m^{-36})^{\frac{2}{3}}(n^{-15})^{\frac{2}{3}}(p^9)^{\frac{2}{3}}$$

and remember the rule must be applied again for each part.
It works like this:
$$(ab^x)^n=a^nb^{xn}$$

As for making all the exponents positive, you don't need to take all the terms into the denominator. For the 3 pronumerals in the question, you only need to take 2 pronumerals and you are allowed to do that.

e.g. $$a^xb^{-x}c^{-y}=\frac{a^x}{b^xc^y}$$

3. Dec 9, 2008

### TayTayDatDude

Re: Exponents

I guess I wrote it wrong, but it's $$(64m^{-36}n^{-15}p^9)^{\frac{2}{3}}$$

4. Dec 9, 2008

### TayTayDatDude

Re: Exponents

Also, would $${(-2x^{-3}y)(-12x^{-4}y^{-2}) / {6xy^{-3}}$$

= $$1/4x^{-6}y^2$$

If so, how do I make the exponents positive??

5. Dec 9, 2008

### Staff: Mentor

Re: Exponents

No.
In the numerator you have (-2)(-12) = 24, and in the denominator you have 6. You should get 24/6 = 4, not 1/4 for the numeric coefficient.

The exponent on x is -3 + (-4) - 1 = -8.
The exponent on y is 2.
To make an exponent positive, replace the exponential factor by its reciprocal. For example, 2x^(-2) = 2*(1/x^2) = 2/(x^2).

6. Dec 9, 2008

### TayTayDatDude

Re: Exponents

I got p^18/16m^72n^30

Is it right? :)

7. Dec 9, 2008

### Staff: Mentor

Re: Exponents

Not even close.
You started with $$(64m^{-36}n^{-15}p^9)^{\frac{2}{3}}$$
That's going to be $$64^{2/3}m^{-36 * 2/3}n^{-15*2/3}p^{9*2/3}$$

For now, simplify 64^(2/3), which is the same as the cube root of 64, squared, and get exponents on the variables that are integers.

After you do that, any variables that have negative exponents can be put in the denominator with positive exponents.

8. Dec 9, 2008

### Mentallic

Re: Exponents

This is exactly the same as what you posted in your first post. The extra brackets for each pronumeral make no difference.

We have $$(64m^{-36})^{\frac{2}{3}}(n^{-15})^{\frac{2}{3}}(p^9)^{\frac{2}{3}}$$

Again, use the rule that $$(a^nb)^x=a^{nx}b^x$$
This must be used for each and every part. See if you can apply this idea to the problem above.

9. Dec 12, 2008

### TayTayDatDude

Re: Exponents

I got

(p^6) / (16m^24)(n^10)

10. Dec 12, 2008

### Дьявол

Re: Exponents

Here is the result, step by step:

$$[(64m^{-36})(n^{-15})(p^9)] ^ {2/3}$$

$$64^{2/3}*m^{-24}*n^{-10}*p^6$$

$$64=2^6$$

$$64^{2/3}=2^{6*2/3}=2^4=16$$

$$16*\frac{1}{m^{24}}*\frac{1}{n^{10}}*p^6$$

$$\frac{16*p^6}{m^{24}*n^{10}}$$

$$(\frac{4*p^3}{m^{12}*n^{5}})^2$$

Last edited: Dec 12, 2008
11. Dec 12, 2008

### TayTayDatDude

Re: Exponents

Uhm, the last statement does not equal the one above it..

12. Dec 12, 2008

### Дьявол

Re: Exponents

Sorry, I got problems with LaTeX code, now it is ok. It should be:

$$(\frac{4*p^3}{m^{12}*n^{5}})^2$$

Last edited: Dec 12, 2008
13. Dec 12, 2008

### Staff: Mentor

Re: Exponents

The last expression (not statement) DOES equal the one above.

14. Dec 12, 2008

### TayTayDatDude

Re: Exponents