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September 18th, 2014, 11:20 PM   #1
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Question Potency laws with fractions?

I know that if you have:

64^(2/3) = (64^(1/3)^2 = 4^2 = 16

But what if you have:

9^(3/2)?

How do you begin to count this? I have one idea but it seems like it's wrong.

9^(3/2) = 9^(1/2)^2 = 3^2 = 9

So how do you count this?
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September 18th, 2014, 11:38 PM   #2
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You have the right general idea, but use instead:

9^(3/2) = (9^(1/2))^3 = 3^3 = 27
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September 19th, 2014, 05:34 AM   #3
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Quote:
Originally Posted by DecoratorFawn82 View Post
I know that if you have:

$$64^{2/3} = (64^{1/3})^2 = 4^2 = 16$$

But what if you have:

$$9^{3/2}?$$

How do you begin to count this? I have one idea but it seems like it's wrong.

$$9^{3/2} = (9^{1/2})^2 = 3^2 = 9$$

So how do you count this?
Basic rule for exponents: $$\huge{(a^m)^n=a^{m\cdot n}}.$$

Applying this to your first problem:

$$\huge{64^\frac{2}{3}}$$

Always write numbers like $4$ as $2^2$, $8$ as $2^3$, $\frac{1}{81}$ as $3^{-4}$, so always with the smallest base possible.

You can write
$\large{64=2^6=4^3}$

Therefore: $$\huge{(2^6)^\frac{2}{3}=2^{6\cdot \frac{2}{3}}=2^4=16}$$

Second problem:
$$\huge{9^\frac{3}{2}=(3^2)^\frac{3}{2}=3^ {2\cdot\frac{3 }{2}}=3^3=27}$$

But we can do the reversed thing as you noticed, that is, using this rule: $$\huge{a^{mn}=(a^m)^n=(a^n)^m}$$

First problem: $$\Large{64^\frac{2}{3}=64^{\frac{1}{3}\cdot 2} =\left(64^\frac{1}{3}\right)^2 = \left(\sqrt[3]{64}\right)^2 =4^2=16.}$$

Second problem: $$\Large{9^\frac{3}{2}=9^{\frac{1}{2}\cdot 3}= \left(9^\frac{1}{2}\right)^3= \left(\sqrt{9}\right)^3=3^3=27.}$$
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September 19th, 2014, 06:14 AM   #4
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Bonus question: what is 27^(5/6)?
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September 19th, 2014, 06:50 AM   #5
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Quote:
Originally Posted by DecoratorFawn82 View Post
I know that if you have:

64^(2/3) = (64^(1/3)^2 = 4^2 = 16

But what if you have:

9^(3/2)?
$\displaystyle 64^{\frac{2}{3}}=(4^3)^{\frac{2}{3}}=4^{3\cdot \frac{2}{3}}=4^2=16
\\\;\\
9^{\frac{3}{2}}=(3^2)^{\frac{3}{2}} + 3^{2\cdot\frac{3}{2}}= 3^3=27$
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September 19th, 2014, 10:54 AM   #6
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(64)^(2/3)
=[(4)^3]^(2/3)
=(4)^[3*(2/3)] [formula: (a^m)^n= (a)^m*n [always keep this general formula to solve exponent problems.]
=(4)^2
=16
Try second one ………………………….
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September 19th, 2014, 11:37 AM   #7
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