Asked by Terrance

System of equations for
-6=Ab^-3
-18=Ab^2
Answered by Steve
Ab^-3 = -6
Ab^2 = -18

divide to get rid of the A's and you have

b^-5 = 1/3
b^5 = 3
b = 3^(1/5)

Now use that to get A
Answered by Terrance
I cant get A it keeps being nonreal

Answered by Steve
Ab^2 = -18
A(3^(1/5))^2 = -18
A(3(2/5)) = -18
A = -18 * 3^(-2/5)
= 18 / 9^(1/5)
= 2 * 9^(4/5)

It would be nice if you showed <u>your</u> work ...
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