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anonymous
 4 years ago
i need help solving this: x^(2/3)2x^(1/3)=15. And the intrsutions say explain and demonstrate how to use substitution to solve the equation. well i don't understand this at all. please help!
anonymous
 4 years ago
i need help solving this: x^(2/3)2x^(1/3)=15. And the intrsutions say explain and demonstrate how to use substitution to solve the equation. well i don't understand this at all. please help!

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0treat x^(2/3) as x^(1/3)^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok put \[z=x^{\frac{1}{3}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then factor as x^22x15=0, and substitute x^1/3 back in for x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then by the laws of exponents \[z^2=(x^{\frac{1}{3}})^2=x^{\frac{2}{3}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so by replacing \[x^{\frac{1}{3}}\] by \[z\] you get an equation that looks like \[z^22z=15\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so the answer should be 5?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there are two answers

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and you can solve this for z by writing \[z^22z15=0\] \[(z5)(z+3)=0\] so \[z=5\] or \[z=3\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that is what i was getting 5 or 3 but then when i try to substitute back in i do not get anything that looks right. i absolutely think that i am doing something wrong still

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but now we have to remember that \[z=x^{\frac{1}{3}}\]so that means that \[x^{\frac{1}{3}}=5\] or \[x^{\frac{1}{3}}=3\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and so \[x=(5)^2=125\] or \[x=(3)^3=27\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so i substitute back in 25 or 127?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you substitute x^(1/3) back in for z

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then cube both sides of each equation to find x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(x^(1/3))^3=5^3 x=125 (x^(1/3))^3=3^3 x=27

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so both are good solutions? im trying to follow sorry, i am just very bad at this

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so both are good solutions? im trying to follow sorry, i am just very bad at this

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes both are good solutions because they are in the domain of the original function.
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