What is the solution to this equation please? #16=(x-2)^(2/3)#

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Answer 1 · 2018-05-28T05:49:46

#x = 66 or x=-62#

I'll assume we're working over the real numbers.

# 16 = (x-2)^{2/3} #

#16^{3/2} = x-2#

#x = 2 + ((16)^{1/2})^{3} #

I interpret the fractional exponents as multivalued; your teacher may have another idea.

#x = 2 + (pm 4 )^{3} #

#x = 2 pm 64 #

#x = 66 or x=-62#

Answer 2 · 2018-05-28T06:57:58

#x = 66 or -62#

Raise both sides to the power of #3/2# to get rid of any power for the #x# term. #" "(2/3 xx3/2 =1)#

#((x-2)^(2/3))^(3/2) =16^(3/2)#

#(x-2) = 16^(3/2)#

#x-2 = (+-sqrt16)^3" "larr x^(p/q) = rootq(x)^p#

#x= (+-4)^3 +2#

This will give two answers:

#x = +64+2 = 66" "or x = -64+2 = -62#