# Solving for Non-Integer Exponents - Help

• January 7th 2010, 11:01 AM
chevywaldo
Solving for Non-Integer Exponents - Help
how do you solve for non-integer exponents.

for example, consider the following

1.5 to the power of .525

1.5^.525

show me some examples please. I'm pretty bad at math.
• January 7th 2010, 11:31 AM
e^(i*pi)
Quote:

Originally Posted by chevywaldo
how do you solve for non-integer exponents.

for example, consider the following

1.5 to the power of .525

1.5^.525

show me some examples please. I'm pretty bad at math.

For that one especially I'd use technology. Let the calculator solve it.

There are some common exponents

$x^{0.5} = \sqrt{x}$

$x^{0.333...} = \sqrt[3]{x}$

$x^{1.5} = \sqrt{x^3}$

In general

$x^{\frac{a}{b}} = \sqrt[b]{x^a}$
• January 7th 2010, 01:12 PM
chevywaldo
Quote:

Originally Posted by e^(i*pi)
For that one especially I'd use technology. Let the calculator solve it.

There are some common exponents

$x^{0.5} = \sqrt{x}$

$x^{0.333...} = \sqrt[3]{x}$

$x^{1.5} = \sqrt{x^3}$

In general

$x^{\frac{a}{b}} = \sqrt[b]{x^a}$

like I said I'm bad at math, so I don't totally understand your response, however I followed your suggestion and allowed my Windows XP calculator to solve for it. The calculator includes a function key for solving problems like these. The key look like this:

x^y

I had to try a few time to learn how this key works. Took me a few tries.

I took my number "x" (which was 1.766) and multiplied it by my "y" value exponent (.526) using the calculator function key and it worked properly.

so it was something like this: 1.766 ^ .526 = 1.34

thanks for your help.
• January 7th 2010, 02:53 PM
Masterthief1324
I know x to the y power is x times itself y number of times. So 4^1.23 means 4 times itself 1.23 times. In terms of a verbal explanation, what does it mean to multiply a number .23 times?

x^(.23*4 + 4)?
• January 7th 2010, 03:10 PM
Quacky
So let's use the example of $4^{0.23}$
This could be written as $4^{\frac{23}{100}}$
Using the: http://www.mathhelpforum.com/math-he...cb3f334f-1.gif rule, we can state that
$4^{\frac{23}{100}} = \sqrt[100]{4^{23}}$

Now let's look at the case of $4^{1.23}$
Using the rule $(x^a)(x^b) = x^{a+b}$
$4^{1.23} = (4^1)(4^{0.23})$
$= 4( \sqrt[100]{4^{23}})$

I'm not sure how to simplify that further, I don't know if it can be.