# Basic nested radical question $\sqrt{y\sqrt{y}}$

First-time asker here. I'm an old guy, going back to college, and I'm in College Algebra. I've run across a problem that I can't match up with the answer that the back of the book (or Wolfram Alpha, or Symbolab) gives.

Here's the problem:

"Use positive exponents to rewrite: $\sqrt{y\sqrt{y}}$ "

The answer is supposed to be $y^{3/4}$, but I can not get it there no matter what I do.

Any help would be very much appreciated!

Barnisinko

$$\sqrt{y\sqrt{y}}=(y\times y^{1/2})^{1/2}=(y^{3/2})^{1/2}=y^{3/4}.$$
You know that $\sqrt{y}=y^{1/2}$ so that: $$y\sqrt{y}=y\cdot y^{1/2}=y^{1+1/2}=y^{3/2}$$ Using again that the square root is the same as a half exponent, we get $$\sqrt{y\sqrt{y}}=\big(y\sqrt{y}\big)^{1/2}=\big(y^{3/2}\big)^{1/2}=y^{(3/2)\cdot (1/2)}=y^{3/4}$$
Rewrite $\sqrt{x} = x^{\frac 1 2}$:
$$\sqrt{y\sqrt{y}} = (y(y)^{\frac 1 2})^{\frac 1 2} = ((y)^{\frac 3 2})^{\frac 1 2} = y^{\frac 3 4}$$