How to cancel out decimal exponents

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• Jul 10th 2012, 08:42 PM
snypeshow
How to cancel out decimal exponents
How would I state this:

10L0.2(2.25L)0.3

in terms of L1 or L?

In previous exercises in my textbook, I've only dealt with a case where L is to the power of 0.5 (L0.5 = sqrt[L])

I know that when you multiply two terms, you add their exponents. So could you restate the question as: 10 x L0.2 x2.250.3 x L0.3 ? I wouldn't know how to go from there though..
• Jul 10th 2012, 09:47 PM
earboth
Re: How to cancel out decimal exponents
Quote:

Originally Posted by snypeshow
How would I state this:

10L0.2(2.25L)0.3

in terms of L1 or L?

In previous exercises in my textbook, I've only dealt with a case where L is to the power of 0.5 (L0.5 = sqrt[L])

You can rearrange a decimal (fraction) into a proper fraction:

$\displaystyle \displaystyle{L^{0.3} = L^{\frac3{10}} = \sqrt[10]{L^3}}$

Quote:

I know that when you multiply two terms, you add their exponents. So could you restate the question as: 10 x L0.2 x2.250.3 x L0.3 ? I wouldn't know how to go from there though..
Your considerations are OK!

In a product the order of the factors can be changed without changing the final result:

$\displaystyle \displaystyle{10 \cdot L^{0.2} \cdot (2.25 L)^{0.3} = 10 \cdot (2.25)^{0.3} \cdot L^{0.2} \cdot L^{0.3}= 10 \cdot (2.25)^{0.3} \cdot L^{\frac12}}$
• Jul 12th 2012, 09:52 AM
snypeshow
Re: How to cancel out decimal exponents
How would I express the final answer in terms of L^1 (or otherwise, just L)?

Could I square the answer? so like: [10 * (2.25)^0.3 * L^0.5]^2

I would get 162.6707657L, would that number still be correct? (I know decimals are evil, but in the actual question, I'd need a decimal number)
• Jul 12th 2012, 10:10 PM
petersonte
Re: How to cancel out decimal exponents
• Jul 12th 2012, 10:33 PM
earboth
Re: How to cancel out decimal exponents
Quote:

Originally Posted by snypeshow
How would I express the final answer in terms of L^1 (or otherwise, just L)?

Could I square the answer? so like: [10 * (2.25)^0.3 * L^0.5]^2

I would get 162.6707657L, would that number still be correct? (I know decimals are evil, but in the actual question, I'd need a decimal number)

I don't understand why you want to square the result of the simplifications(?).

In your 1st post you wanted to transform a term. Squaring this term will change it's output if you plug in values of L.

So the only possible way to get a single L is - in my opinion - :

$\displaystyle 10 \cdot (2.25)^{0.3} \cdot L^{\frac12} = 10 \cdot \left(\frac94\right)^{\frac3{10}} \cdot \sqrt{L}$

... but I'm not sure if it is this what you are looking for(?)
• Jul 12th 2012, 10:54 PM
snypeshow
Re: How to cancel out decimal exponents
Yeah, that's what I wanted!! Thanks!