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When I try to solve the equation, I get $x^{1/4}$ since the two roots share the same index and exponent, which means that I ought to be able to multiply across.

Instead, the solution in the textbook is as follows.

$$\left (xx^{1/2} \right )^{1/2}=\left ( x^{3/2} \right )^{1/2}=x^{3/4}$$

Where did $x^{3/2}$ come from? I can't figure out how I get a 3 from any of that. And why is my original intuition incorrect?

YiFan Tey
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Nathaniel
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3 Answers3

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The notation is $\sqrt{x\sqrt x}= m$

Then $m^2 = x\sqrt x$.

So $(m^2)^2 = (x\sqrt x)^2 = x^2*(\sqrt x)^2 = x^2*x = x^3$

So $m^4 = x^3$ and so

$m = \sqrt[4]{x^3} = x^{\frac 34}$

....

The book did it this way:

$\sqrt{x\sqrt x} = (x\cdot x^{\frac 12})^{\frac 12}=$

$(x^1\cdot x^{\frac 12})^{\frac 12} =$

$(x^{1+ \frac 12})^{\frac 12} =$

$(x^{\frac 32})^{\frac 12} =$ (this is where the $\frac 32$ came from. $1+\frac 12 = \frac 32$).

$x^{\frac 32\cdot \frac 12} = x^{\frac 34}$.

......

Note: $\sqrt{\sqrt x} = x^{\frac 14}$ is where your intuitions comes from. But $\sqrt{x \sqrt x}$ has an extra $x$ in it. But there is more than one way to do things: $\sqrt{x \sqrt{x}}=\sqrt{\sqrt{x^2}\sqrt x} = \sqrt{\sqrt{x^2*x}} = \sqrt{\sqrt{x^3}} = (x^3)^{\frac 14} = x^{\frac 34}$

fleablood
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  • So, why is $[\left ( {x}\cdot {x}^{1/2} \right )^{1/2}]$ not $\left ( {x}^{1/2}\cdot {x}^{1/2} \right )$? – Nathaniel Oct 26 '19 at 03:40
  • Do both x's not share the same exponent and index? And why does that switch to addition? – Nathaniel Oct 26 '19 at 03:40
  • @Nathaniel You have to distribute the exponent to both terms. $$(x\cdot x^{1/2})^{1/2}=x^{1/2}\cdot x^{(1/2)(1/2)}=x^{1/2}\cdot x^{1/4}=x^{(1/2)+(1/4)}=x^{3/4}.$$ – YiFan Tey Oct 26 '19 at 03:47
  • I'm completely fine on that part. But I cannot understand why the equation is not written $\sqrt{x}\cdot \sqrt{x} = x^{1/2}\cdot x^{1/2}$.

    I'm sorry for being obtuse, but I cannot wrap my head around why that first x root is not the exact same thing as the second root. There should be an invisible 2 above the first root serving as the index, and an invisible 1 exponent above the x. And the same should be true for the second x root. Then we should be multiplying them.

    Why is that not the case?If a root is within another root, do they have some other property?Are we not multiplying?

     – Nathaniel Oct 26 '19 at 03:59
  • @Nathaniel I think what you're missing is that the second $\sqrt{x}$ is under another layer of square root as well. So $\sqrt{x\sqrt{x}}$ is not the same thing as $\sqrt{x}\sqrt{x}$ although they look almost identical; instead, $\sqrt{x\sqrt{x}}$ is the same as $\sqrt{x}\sqrt{\sqrt{x}}$. The bigger square root also encapsulates the second term. – YiFan Tey Oct 26 '19 at 04:26
  • $(a\cdot b)^{\frac 12} = a^{\frac 12}b^{\frac 12}$ and if $a = x$ and $b = \sqrt{x}$ then $(x\cdot x^{\frac 12})^{\frac 12} = (x)^{\frac 12}(x^{\frac 12})^{\frac 12}$. – fleablood Oct 26 '19 at 05:17
  • I can't wrap me head around why you think the first term under the root, which is $x$, should be the same thing as the second term under the root wihich is $\sqrt x$. As you know, $x$ and $\sqrt{x}$ are not the same thing. So we are you treating them as though they are. – fleablood Oct 26 '19 at 05:20
  • "Do both x's not share the same exponent and index?" No. They absolutely do not. And now I think I see your confusion. You are confusing $x\cdot x^{\frac 12}$ with $(x \cdot x)^{\frac 12}$. Those are two entirely differenth things. When you have an $ab^k$ the $k$ only applies to the $b$ it does not apply to the $a$. So when you have $x \cdot x^{\frac 12}$ then $\frac 12$ only applies to the second $x$. The first $x$ is a $x^1$. $x\cdot x^{\frac 12} = x^1\cdot x^{\frac 12}= x^{1+\frac 12}$. – fleablood Oct 26 '19 at 05:28
  • "If a root is within another root, do they have some other property?Are we not multiplying? " We are doing that. $\sqrt[2]{x^1\cdot x^{\frac 12}} = x^{1\cdot \frac 12}\cdot x^{\frac 12 \cdot \frac 12}$. That's multiplying. – fleablood Oct 26 '19 at 05:31
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We have

$$xx^{1/2}=x^1x^{1/2}=x^{1+1/2}=x^{3/2} $$

therefore

$$\sqrt{x\sqrt{x}}=\left(xx^{1/2}\right)^{1/2}=\left(x^1x^{1/2}\right)^{1/2}=\left(x^{3/2}\right)^{1/2}=x^{3/4}$$

Axion004
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What you're thinking of is $\sqrt{\sqrt{x}}=x^{1/4}.$ However, by exponent laws, the expression you have is actually equal to $\sqrt{x\sqrt{x}}=\sqrt{x^{3/2}}=x^{3/4}$

Simon Fraser
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