SOLUTION: Dear Algebra II tutor, I am having trouble solving and checking thje extraneous solution of the following problem: (2x+1)1/3 = 3 note: the 1/3 is an exponent. I tried it out a

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Dear Algebra II tutor, I am having trouble solving and checking thje extraneous solution of the following problem: (2x+1)1/3 = 3 note: the 1/3 is an exponent. I tried it out a      Log On


   

Question 121080: Dear Algebra II tutor,
I am having trouble solving and checking thje extraneous solution of the following problem:
(2x+1)1/3 = 3
note: the 1/3 is an exponent.
I tried it out and got x=4 here is my work:
(2x+1)1/3(3) = 3(3)
2x+1=9
2x=8
x=4
I would very much appriate any help that you may offer, thank you for your time and effort.
Justin
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
What you have is %282x%2B1%29%5E%281%2F3%29=3, which is the same thing as root%283%2C2x%2B1%29=3, so rather than multiplying both sides of the equation by 3, you have to raise both sides of the equation to the 3rd power, thus:

%28%282x%2B1%29%5E%281%2F3%29%29%5E3=3%5E3

Now apply the rule: %28x%5Em%29%5En=x%5Emn

and since 3%5E3=27,

2x%2B1=27
2x=26
x=13

Check:
Does %282%2813%29%2B1%29%5E%281%2F3%29=3?

%282%2813%29%2B1%29%5E%281%2F3%29=%2827%29%5E%281%2F3%29=root%283%2C27%29=3 checks!