How do you find all real number solutions to #root3(2x+3)-1=0#?

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Answer 1 · 2016-11-13T22:59:33

#root(3)(2x+ 3) = 1#

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

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

#2x + 3 = 1#

#2x = -2#

#x = -1#

Hopefully this helps!