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anonymous
 one year ago
(x^2y^4x^3)^2 First remove the parenthesis now we use the ab = 1/ab property x^2 = 1/x^2 now we have a positive exponent x^2. Next y^4 = 1/y^4 now we have three positive exponents now we need to make the negative 2 positive. 2= 1/2
now we have x^2y^4x^3 add the like terms x^2+ x^3 = x^5
(x^5y^4)^2
anonymous
 one year ago
(x^2y^4x^3)^2 First remove the parenthesis now we use the ab = 1/ab property x^2 = 1/x^2 now we have a positive exponent x^2. Next y^4 = 1/y^4 now we have three positive exponents now we need to make the negative 2 positive. 2= 1/2 now we have x^2y^4x^3 add the like terms x^2+ x^3 = x^5 (x^5y^4)^2

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can you help me with this @misssunshinexxoxo

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\LARGE\color{black}{ \displaystyle \large \left(x^{2}~y^{4}~x^3\right)^{2} }\) like this?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah. Brb five minutes or less sorry making lunch.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Ok, I will tell you another property..... \(\displaystyle \Large {\rm w}^\color{red}{\rm \LARGE a}\times{\rm w}^\color{blue}{\rm \LARGE b}={\rm w}^{\color{red}{\rm \LARGE a}+\color{blue}{\rm \LARGE b}}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1this property, can be applied to the terms with an x. lets do a little replacement, just to visualize it better: \(\LARGE\color{black}{ \displaystyle \large \left(x^{2}~\cdot~y^{4}~\cdot~x^3\right)^{2} }\) \(\LARGE\color{black}{ \displaystyle \large \left(x^{2}~\cdot~x^3~\cdot~y^{4}\right)^{2} }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1so, you can apply the property I told you just now to the first 2 terms in this expression.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1ok, take your time to read, and apply the property i have provided...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So we don't change the terms into positive exponents?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1we will do that later, but for now we just want to simplify inside the parenthesis (using this new property i told you).....

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\LARGE\color{black}{ \displaystyle \large \left(x^{2}~\cdot~x^3~\cdot~y^{4}\right)^{2} }\) (this is your expression  the order of how you multiply doesn't matter)  Knowing that: \(\displaystyle \Large {\rm w}^\color{red}{\rm \LARGE a}\times{\rm w}^\color{blue}{\rm \LARGE b}={\rm w}^{\color{red}{\rm \LARGE a}+\color{blue}{\rm \LARGE b}}\) then, \(\displaystyle \Large {\rm x}^\color{red}{\rm \LARGE 2}\times{\rm w}^\color{blue}{\rm \LARGE 3}={\rm x}^{?}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0By the bases being different i can't add them together right. so i guess x would be still x^2

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1you can apply it to the " x\(^{2}\) • x\(^3\) " part (where the base is same, the base is x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we would just have x^1 which we don't count so we just have x.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry thought i already enter it forgot i had to click on post. >~< @SolomonZelman

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1yes, x^1 is correct.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1And x^1 is the same exact thing as x (besides the fact that you waste extra space and time when writing ^1 there)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1we had: \(\LARGE\color{black}{ \displaystyle \large \left(x^{2}~\cdot~x^3~\cdot~y^{4}\right)^{2} }\) our new expression (that is equivalent to the previous one, of course) is: \(\LARGE\color{black}{ \displaystyle \large \left(x~\cdot~y^{4}\right)^{2} }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1now, apply the property of \(a^{b}\)=\(\dfrac{1}{a^b}\) to the " y\(^{4}\) " part

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay y^4 =1/y^4 @SolomonZelman

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So how do i solve for 2 then?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1our expression: \( \LARGE\color{black}{ \displaystyle \large \left(x~\cdot~y^{4}\right)^{2} }\) is going to become: \( \LARGE\color{black}{ \displaystyle \large \left(x~\cdot~\dfrac{1}{y^{4}}\right)^{2} }\) \( \LARGE\color{black}{ \displaystyle \large \left(\dfrac{x}{y^{4}}\right)^{2} }\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so we don't change two?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1I will give you another propery (which can is derived from \(a^{b}\)=1 /\(a^b\)) the property is: \(\color{black}{ \displaystyle \large \left(\dfrac{\color{red}{\rm a}}{\color{blue}{\rm b}}\right)^{\Largew} = \left(\dfrac{\color{blue}{\rm b}}{\color{red}{\rm a}}\right)^{\Large w} }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Text  correct: `which can *be* derived...`

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Knowing that: \(\color{black}{ \displaystyle \large \left(\dfrac{\color{red}{\rm a}}{\color{blue}{\rm b}}\right)^{\Largew} = \left(\dfrac{\color{blue}{\rm b}}{\color{red}{\rm a}}\right)^{\Large w} }\) then, therefore: \(\color{black}{ \displaystyle \large \left(\dfrac{\color{red}{ x}}{\color{blue}{ y^4}}\right)^{\Large2} = \left(\dfrac{\color{blue}{y^4}}{\color{red}{x}}\right)^{\Large 2} }\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah just make the reciprocal and it turn the negative into a positive.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for your help again. I appreciate it. @SolomonZelman

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\color{black}{ \displaystyle \large \left(\dfrac{\color{red}{ x}}{\color{blue}{ y^4}}\right)^{\Large2} = \left(\dfrac{\color{blue}{y^4}}{\color{red}{x}}\right)^{\Large 2}=\dfrac{\left(\color{blue}{y^4}\right)^2}{\left(\color{red}{x}\right)^2} }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1yes, this one more step...

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\((\)y\(^4\)\()\)\(\large^{^2}\) = y\(^{4~\times~ 4}\) = y\(^?\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1yes, y\(^{16}\)/x\(^2\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1everything makes sense?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah so far do i divide by x or i'm just writing it as y^16 over x^2

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1you had \(\color{black}{ \displaystyle \large \left(\dfrac{\color{blue}{y^4}}{\color{red}{x}}\right)^{\Large 2}=\dfrac{\left(\color{blue}{y^4}\right)^2}{\left(\color{red}{x}\right)^2} }\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we multiply the exponent y^4 with 2 and we get 16/x^2 Yeah it makes sense :)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1applying the rule: \(\color{black}{ \displaystyle \large \left(\dfrac{\color{blue}{\rm A}}{\color{red}{\rm B}}\right)^{\Large w}=\dfrac{\left(\color{blue}{\rm A}\right)^w}{\left(\color{red}{\rm B}\right)^w} }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1and then after we applied this rule above, we simplify on the top.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Which gives the result of: y\(^{16}\)/x\(^2\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1and that is it. can't do anything with this....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright thanks @SolomonZelman
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