Question with exponents!
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- calculusxy

Question with exponents!
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- schrodinger

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- calculusxy

\[\huge \frac{ (-xy^4)^{-4} }{ -2x^3 \times x^2y^3 }\]

- Jhannybean

\[(-2x^3)(x^2y^2) \implies -2 \cdot x^{3+2} \cdot y^2 =~?\]

- calculusxy

i got the answer of \[\frac{ 1 }{ 2x^9y^19 }\] but the answer key says that it's negative.. i don't know why

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## More answers

- Jhannybean

Alright let's check each step.

- Jhannybean

what would you get for :
\(\color{#0cbb34}{\text{Originally Posted by}}\) @Jhannybean
\[\large (-2x^3)(x^2y^2) \implies -2 \cdot x^{3+2} \cdot y^2 =~?\]
\(\color{#0cbb34}{\text{End of Quote}}\)

- calculusxy

\[\large -2x^5y^2\]

- Jhannybean

Alright, and when you distribute the \(-4\) power into every term in the ( ), you expand it like so: \[\large (-xy^4)^{-4} \implies (-x)^{-4} \cdot (y^4)^{-4} = ~?\]

- calculusxy

\[\large -x^{-4} \times y^0\]

- calculusxy

which is basically like \[\large -x^{-4}\]

- calculusxy

\[\large \times \]
do you mean this symbol? if so, it's multiplication.

- Jhannybean

be careful, you wouldnt be adding the exponents here, you'd be multiplying them. \[\large (y^4)^{-4} = y^{4( -4)}\qquad (y^4)^{-4} \ne y^{4-4}\]

- calculusxy

oh yes

- calculusxy

i still don't understand how you got the negative...

- calculusxy

yeah or whose ever work that was posted in the 3 parts

- just_one_last_goodbye

pretty fast eh? ^_^ thinking about making it a tool on OS as an extention

- Jhannybean

defeats the purpose of Openstudy then.

- calculusxy

can you please help me understand why it is a negative?

- Jhannybean

Ok cotinuing from where we left off.

- Jhannybean

I just realized theres an easier way to simplify the numerator, check this out

- Jhannybean

Instead of expanding our numerator first, we can basically turn our entire term WITH the negative exponent into a positive one. \[\large (-xy^4)^{-4} \implies \dfrac{1}{(-xy^4)^4}\] you see how the negatives in the numerator would go away?

- Jhannybean

So now we expand our numerator, and we get: \[\large \frac{1}{x^4y^{16}}\]

- Jhannybean

Now we put it over our denominator: \[\large \dfrac{\dfrac{1}{x^4y^{16}}}{-2x^5y^3}\qquad \implies \qquad \dfrac{1}{-2x^{4+5}y^{16+3} } \]

- Jhannybean

Now whether we have the negative in the numerator OR denominator, it doesnt matter, the fraction altogether is STILL negative.

- Jhannybean

Do you follow, @calculusxy ?

- calculusxy

i am still trying to grasp this.

- calculusxy

if we have like complex fractions, then we would have to add the exponents n put it over one?

- Jhannybean

The only reason we did that, was because we treated \((-xy^4)\) as like a variable... let' say all that junk \(=a\). And then... a is raised to a negative power, and how would we turn that power positive? by putting it over 1. \(a^{-1} \iff \dfrac{1}{a}\)

- Jhannybean

So if we have a FRACTION in the numerator, and we're DIVIDING it by a term or terms in the denominator, we follow this rule: \[\dfrac{\dfrac{a}{b}}{c} \implies \frac{a}{bc}\]

- calculusxy

okay...

- calculusxy

i have one other question..

- calculusxy

thanks for helping me understand that

- Jhannybean

Which art are you stuck on? WE'll go from there.

- Jhannybean

part*

- calculusxy

so this time i am getting a negative, but it says that it's a positive.
\[\large -\frac{ ba^0 \times -a^3b^3 }{ (ab^0)^{-4} }\]

- calculusxy

so i got \[\large -a^7b^4\] but the answer key simply says \[\large a^7b^4\]

- calculusxy

- calculusxy

- calculusxy

@jim_thompson5910 it's the second expression w/ exponents.

- jim_thompson5910

##### 1 Attachment

- calculusxy

oh okay thx !

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