0

I'm trying to isolate $y$. I have a constant times a negative one? Do I ignore the negative and leave is as a constant? Here's what I'm working with...

$$|(6-2(y^3))| = ke^{-3\times2}$$

For a positive portion: $6-2(y^3) = ke^{-3\times-2}$

For negative portion: $6-2(y^3) = -ke^{-3\times-2}$

Does the negative portion just simplify to $ke^{-3\times-2}$ ? I don't understand the rules of constants...

Martín-Blas Pérez Pinilla
  • 41,546
  • 4
  • 46
  • 89
Alex M
  • 11

1 Answers1

1

You have to take into account the sign of $k$. In the positive case, you have $K>0$ since exponential is always positive, in the other you have $-k <0$ which is again $k>0$. You can simplify this to just $k e^{-3x-2}$ as you said, but then you have $k<0$.

user26977
  • 1,118
  • ok. I think I understand. So let's say I simplify to -2(y^3) = ke^(-3x^2)-6. Would that mean y^3 = ke^(-3x^2)+3 ? – Alex M Apr 08 '16 at 21:48
  • yes, but still you have 2 cases, if $y^3 <3$ then $k>0$ and if $y^3\geq 3$ then $k\leq 0$ – user26977 Apr 08 '16 at 21:52