Simplify the following problem with exponents

[TayTayDatDude]

Homework Statement

Simplify. Express final answers with positive exponents.

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

Homework Equations

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

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.

 

[Mentallic]

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

 

[TayTayDatDude]

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

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

 

[TayTayDatDude]

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

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??

 

[Mark44]

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??

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).

 

[TayTayDatDude]

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

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

Is it right? :)

 

[Mark44]

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.

 

[Mentallic]

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

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.

 

[TayTayDatDude]

I got

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

 

[Дьявол]

Here is the result, step by step:

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

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

$$64=2^6$$

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

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

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

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

 

[TayTayDatDude]

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

 

[Дьявол]

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

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

 

[Mark44]

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

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

 

[TayTayDatDude]

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

Yes, now it does. Please care to read his edit.