OpenStudy (anonymous):

does anyone understand this finding the real solution? the 3 and 2 are exponets ???x^3/^2 -27=0

7 years ago

Hm. What do you mean? The real solution of each is the appropriate root. e.g.: $$x^2 - 27 = 0$$ $$x^2 = 27$$ $$x = \pm \sqrt{27}$$

7 years ago

Ah, sorry, misunderstood. You have $$x^{\frac{3}{2}}$$ In that case the process is similar, but you have to remember that $$\sqrt{x}$$ is the same as $$x^2$$, so $$x^{\frac{3}{2}}$$ is really the same as $$\sqrt{x^3}$$. So, you end up with: $$\sqrt{x^3} = 27$$ You can reverse the square root by squaring and the cube by taking the cubed root.

7 years ago
I meant $$\sqrt{x}$$ is the same as $$x^{\frac{1}{2}}$$.