Basic nested radical question √y√y \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: √y√y \sqrt{y\sqrt{y}} ”

The answer is supposed to be y3/4 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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3 Answers
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You have

√y√y=(y×y1/2)1/2=(y3/2)1/2=y3/4.\sqrt{y\sqrt{y}}=(y\times y^{1/2})^{1/2}=(y^{3/2})^{1/2}=y^{3/4}.

  

 

Thank you very much! This was very helpful!
– Barnisinko
Oct 20 at 20:23

You know that √y=y1/2\sqrt{y}=y^{1/2} so that:
y√y=y⋅y1/2=y1+1/2=y3/2y\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
√y√y=(y√y)1/2=(y3/2)1/2=y(3/2)⋅(1/2)=y3/4\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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Thanks so much for your help!
– Barnisinko
Oct 20 at 20:22

Rewrite √x=x12\sqrt{x} = x^{\frac 1 2}:

√y√y=(y(y)12)12=((y)32)12=y34 \sqrt{y\sqrt{y}} = (y(y)^{\frac 1 2})^{\frac 1 2} = ((y)^{\frac 3 2})^{\frac 1 2} = y^{\frac 3 4}

  

 

Thanks so much, I really appreciate it!
– Barnisinko
Oct 20 at 20:23

  

 

You don’t need to say thank you, just upvote the answer if it was helpful.
– adjan
Oct 20 at 20:25