Why does $\sqrt{x\sqrt{x}}$ equal $x^{3/4}$ instead of $x^{1/4}$?

Question

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?

Answer 1

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: $\sqrt{x \sqrt{x}}=\sqrt{\sqrt{x^2}\sqrt x} = \sqrt{\sqrt{x^2*x}} = \sqrt{\sqrt{x^3}} = (x^3)^{\frac 14} = x^{\frac 34}$

Answer 2

We have

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

therefore

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

Answer 3

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