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September 18th, 2014, 11:20 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, 11:38 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs 
You have the right general idea, but use instead: 9^(3/2) = (9^(1/2))^3 = 3^3 = 27 
September 19th, 2014, 05:34 AM  #3  
Newbie Joined: Sep 2014 From: . Posts: 15 Thanks: 2  Quote:
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: $$\Large{9^\frac{3}{2}=9^{\frac{1}{2}\cdot 3}= \left(9^\frac{1}{2}\right)^3= \left(\sqrt{9}\right)^3=3^3=27.}$$  
September 19th, 2014, 06:14 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, 06:50 AM  #5 
Senior Member Joined: Apr 2014 From: Europa Posts: 575 Thanks: 176  
September 19th, 2014, 10:54 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, 11:37 AM  #7 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs 
Anyone else want to pile on? 

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