# Math question

1. Jan 22, 2005

how is (8x^2)/(sqrt(x)) the same as 8x^(3/2)?

let's see, i can convert that to (8x^2)/(x^(1/2))

but i dont see how it is the same as 8x^(3/2)... can someone lend me a hand?

2. Jan 22, 2005

### cepheid

Staff Emeritus
It's simply the laws of exponents: when you divide two powers with the same base, you simply subtract the exponents:

x2 / x1/2

= x 2 - 1/2 = x 4/2 - 1/2 = x3/2

In general, when you multiply, add the exponents. When you divide, subract the exponents, When you raise the power to another power, multiply the exponents. Notice that for positive integer exponents, these laws are very easy to understand: x cubed times x squared is three x's multiplied together times two x's multiplied together. How many x's are there multiplied together in total at the end? Five. And if you had added the exponents: 3+2 =5.

Even in this example, it's pretty easy to understand: consider this:

$$(\sqrt{x})^2 = x$$

Therefore: $$x^2 = (\sqrt{x})^2 \cdot (\sqrt{x})^2 = (\sqrt{x})(\sqrt{x})(\sqrt{x})(\sqrt{x})$$

Therefore: $$x^2 / x^{1/2} = \frac{(\sqrt{x})(\sqrt{x})(\sqrt{x})(\sqrt{x})}{\sqrt{x}} = (\sqrt{x})(\sqrt{x})(\sqrt{x}) = (x^{1/2})^3 = x^{3/2}$$

As far as I know, the laws hold for any real exponents, positive or negative.

Last edited: Jan 22, 2005
3. Jan 22, 2005

### fourier jr

it doesn't hold just for any real exponent but any complex one also. i think complex exponentiation is defined as $$c^z = exp(clogz)$$ for complex c, z.

4. Jan 23, 2005

### Hurkyl

Staff Emeritus
This property of exponents work when the base is positive and the exponents are real.

For negative or complex bases, or complex exponents, this property fails fairly often -- you should generally try not to apply it to them.