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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

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You have

$$\sqrt{y\sqrt{y}}=(y\times y^{1/2})^{1/2}=(y^{3/2})^{1/2}=y^{3/4}.$$

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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 $$\sqrt{y\sqrt{y}}=\big(y\sqrt{y}\big)^{1/2}=\big(y^{3/2}\big)^{1/2}=y^{(3/2)\cdot (1/2)}=y^{3/4}$$

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Rewrite $\sqrt{x} = x^{\frac 1 2}$:

$$ \sqrt{y\sqrt{y}} = (y(y)^{\frac 1 2})^{\frac 1 2} = ((y)^{\frac 3 2})^{\frac 1 2} = y^{\frac 3 4} $$

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  • $\begingroup$ Thanks so much, I really appreciate it! $\endgroup$ – Barnisinko Oct 20 '16 at 20:23
  • $\begingroup$ You don't need to say thank you, just upvote the answer if it was helpful. $\endgroup$ – Adrian Oct 20 '16 at 20:25

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