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 September 18th, 2014, 10:20 PM #1 Member   Joined: Sep 2014 From: Sweden Posts: 94 Thanks: 0 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? September 18th, 2014, 10:38 PM #2 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs You have the right general idea, but use instead: 9^(3/2) = (9^(1/2))^3 = 3^3 = 27 Thanks from topsquark September 19th, 2014, 04:34 AM   #3
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 Originally Posted by DecoratorFawn82 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: $$\Large{64^\frac{2}{3}=64^{\frac{1}{3}\cdot 2} =\left(64^\frac{1}{3}\right)^2 = \left(\sqrt{64}\right)^2 =4^2=16.}$$

Second problem: $$\Large{9^\frac{3}{2}=9^{\frac{1}{2}\cdot 3}= \left(9^\frac{1}{2}\right)^3= \left(\sqrt{9}\right)^3=3^3=27.}$$ September 19th, 2014, 05:14 AM #4 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Bonus question: what is 27^(5/6)? September 19th, 2014, 05:50 AM   #5
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Quote:
 Originally Posted by DecoratorFawn82 I know that if you have: 64^(2/3) = (64^(1/3)^2 = 4^2 = 16 But what if you have: 9^(3/2)?
$\displaystyle 64^{\frac{2}{3}}=(4^3)^{\frac{2}{3}}=4^{3\cdot \frac{2}{3}}=4^2=16 \\\;\\ 9^{\frac{3}{2}}=(3^2)^{\frac{3}{2}} + 3^{2\cdot\frac{3}{2}}= 3^3=27$ September 19th, 2014, 09:54 AM #6 Newbie   Joined: Sep 2014 From: hyderabad Posts: 9 Thanks: 1 (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 …………………………. September 19th, 2014, 10:37 AM #7 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Anyone else want to pile on?  Tags fractions, laws, potency rules of potency math

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