Rational expressions with powers to an exponent.

SourPatchParent

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I am curious about if I am doing this right for I am trying to teach it to myself without any teacher guidance unfortunately..I could be way off but I am not currently seeing any other way.

The questions I have for this example is:

(x^{7a})^{a-b}

/ (x^{3a})^{a-2b}



1/x^{4b}

(because I had to have it as a positive exponent)


Any advice or help would be greatly appreciated!

Mel
 
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I am curious about if I am doing this right for I am trying to teach it to myself without any teacher guidance unfortunately..The questions I have for this example is:
(x^7a)^a-b / (x^3a)^a-2b
Please please learn to post what you mean.

I think you mean \(\displaystyle (x^{7a})^{a-b}\) but you wrote \(\displaystyle (x^7a)^a-b\).

Write (x^{7a})^{a-b}.

Then \(\displaystyle \displaystyle{(x^{7a})^{a-b}=x^{7a^2-7ab}}\).
 
(x^{7a})^{a-b}

/ (x^{3a})^{a-2b}

1/x^{4b}

(because I had to have it as a positive exponent)
Are the first two lines the "question", with the third line being your "answer"? If so, is the expression meant to be as follows?

. . . . .\(\displaystyle \dfrac{\left(x^{7a}\right)^{a\, -\, b}}{\left(x^{3a}\right)^{a\, -\, 2b}}\)

(For formatting advice, try here.)

Were the instructions something like "simplify, using only positive exponents"? If so, by what steps did you arrive at your answer?

Please be complete. Thank you! ;)
 
Are the first two lines the "question", with the third line being your "answer"? If so, is the expression meant to be as follows?

. . . . .\(\displaystyle \dfrac{\left(x^{7a}\right)^{a\, -\, b}}{\left(x^{3a}\right)^{a\, -\, 2b}}\)

(For formatting advice, try here.)

Were the instructions something like "simplify, using only positive exponents"? If so, by what steps did you arrive at your answer?

Please be complete. Thank you! ;)

Sorry I am new to the site, and I am taking grade 12 advanced functions because I got into a university program with the conditional offer that I complete it by September. I know the laws of the exponents but i get a bit lost when I have to combine a few of them to get an answer. And yes, you are correct--that is what I was "trying" to express.

\(\displaystyle \dfrac{\left(x^{7a}\right)^{a\, -\, b}}{\left(x^{3a}\right)^{a\, -\, 2b}}\)

is the question.

x^{7a^2} - ^{7ab} / x^{3a^2} - ^{6ab}

x^{4-ab}

I have yet to master how to type these sorts of equations out in text but I hope you can bear with me until I do.
 
Sorry I am new to the site, and I am taking grade 12 advanced functions because I got into a university program with the conditional offer that I complete it by September. I know the laws of the exponents but i get a bit lost when I have to combine a few of them to get an answer. And yes, you are correct--that is what I was "trying" to express.

\(\displaystyle \dfrac{\left(x^{7a}\right)^{a\, -\, b}}{\left(x^{3a}\right)^{a\, -\, 2b}}\)

is the question.

x^{7a^2} - ^{7ab} / x^{3a^2} - ^{6ab}

x^{4-ab}

I have yet to master how to type these sorts of equations out in text but I hope you can bear with me until I do.
So what is your answer?
 
\(\displaystyle \dfrac{\left(x^{7a}\right)^{a\, -\, b}}{\left(x^{3a}\right)^{a\, -\, 2b}}\)

is the question.

x^{7a^2} - ^{7ab} / x^{3a^2} - ^{6ab}

x^{4-ab}
.
Here are similar type problems
(x^a)^(3b-c) = x^(3ab-ac)
[x^d]/[x^e]=x^(d-e)
[(x^c)^(3a-2b)/[(x^c)^(a-b)] = [x^(3ac-2bc)]/[x^(ac-bc)]= x^(2ac-bc)
 
Here are similar type problems
(x^a)^(3b-c) = x^(3ab-ac)
[x^d]/[x^e]=x^(d-e)
[(x^c)^(3a-2b)/[(x^c)^(a-b)] = [x^(3ac-2bc)]/[x^(ac-bc)]= x^(2ac-bc)

Can the third problem be simplified further? Since "x" is a common base can I subtract ac from 3 ac and bc from 2bc to give me x^(2ac-bc)?
 
Now you confused me. You compute the top power minus the bottom power.

What the heck is the fuss all about?
\(\displaystyle \dfrac{\left(x^{7a}\right)^{a-b}}{\left(x^{3a}\right)^{a-2b}}=\dfrac{\left(x^{7a^2-7ab}\right)}{\left(x^{3a^2-6ab}\right)}=\left(x^{4a^2-ab}\right)\)
 
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