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I'm doing some revision on indices and surds. How do you simplify

$(xy^2)^p\sqrt{x^q} $

Bit confused because my textbook says the answer is

$x^{p+q/2}y^{2p}$

I can understand simplifying but only when it's the same base. I'm confused with this specific question - step-by-step help would be much appreciated!

[[Edit: changed $x^{p+1/2q}$ to $x^{p + q/2}$.]]

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  • $\begingroup$ The question isn't really clear.Use MathJax for math formatting- math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Naive Jul 20 '17 at 15:09
  • $\begingroup$ $(xy^2)^p\sqrt{x^q}=x^p(y^2)^p (x^q)^{1/2}=x^py^{2p}x^{q/2}=x^{p+q/2}y^2p$ $\endgroup$ – Surb Jul 20 '17 at 15:17
  • $\begingroup$ To improve future posts, include your attempt rather than just describe what you know. Also, mathjax. $\endgroup$ – Siong Thye Goh Jul 20 '17 at 15:34
  • $\begingroup$ Why is no answer accepted? $\endgroup$ – Ramanujan Jan 17 '19 at 17:54
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Hint:

Great you can handle when things are in the same base.

Simplify $$x^p \sqrt{x^q}$$

and simplify $$(y^2)^p$$

Remark: Note that

$$(1/2)q = q/2$$

but $$(1/2)q \neq \frac{1}{2q}$$

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$$(x\cdot y^2)^p\cdot\sqrt{x^q}=x^p\cdot(y^2)^p \cdot(x^q)^{1/2}=x^p\cdot y^{2p}\cdot x^{q/2}\\=x^p\cdot x^{q/2}\cdot y^{2p}=(x^p\cdot x^{q/2})\cdot y^{2p}=x^{p+q/2}\cdot y^{2p}$$

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You need parentheses to clarify your expression, and you seem to have dropped a "^"- you mean (xy^2)^p sqrt(x^q). Apply the "rules of exponents": (xy^2)^p= x^p (y^2)^p= x^p y^{2p}. Also sqrt(x^q)= (x^q)^(1/2)= x^(q/2).

So (xy^2)^p sqrt(x^q)= (x^p y^(2p))(x^(q/2))= x^(p+ q/2) y^{2p}

Also "q/2" is NOT the same as "1/2q"!

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  • $\begingroup$ Is there an alternative answer?? I'm just wondering why my textbook says x^{p+1/2q} $\endgroup$ – secretagentsky Jul 20 '17 at 15:34
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    $\begingroup$ $(1/2)q = q/2$, $(1/2)q \neq \frac1{2q}$. $\endgroup$ – Siong Thye Goh Jul 20 '17 at 15:36
  • $\begingroup$ @SiongThyeGoh this seems indeed to have been the confusion of OP. You may want to add this in your answer. $\endgroup$ – Surb Jul 20 '17 at 16:26
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    $\begingroup$ Sure, thanks for the suggestion. $\endgroup$ – Siong Thye Goh Jul 20 '17 at 16:41

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