How do you solve #4x^(2/3)-6=10#?

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Answer 1 · 2016-02-28T13:48:32

#x=(5^(3/2))/(2^(3/2)#

#color(blue)(4x^(2/3)-6=10#

#rarrx^(2/3)=10/4#

#rarrx^(2/3)=5/2#

Take root #2/3# both sides

#rarrx=root(2/3)(5/2)#

Invert and multiply

#rarrx=(5/2)^(3/2)#

Remember that #(a/b)^x=a^x/b^x#

So

#rArrcolor(green)(x=(5^(3/2))/(2^(3/2))~~39.25...#

Answer 2 · 2016-02-28T13:52:52

#x= +-8#

#4x^(2/3)-6=10#

#rarr 4x^(2/3) = 16#

#rarr x^(2/3)=4#

#rarr x^(1/3) = +-2#

#rarr x= (+-2)^3 = +-8#