My Math Forum Fractional Exponents

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 March 10th, 2016, 02:33 AM #1 Newbie   Joined: Mar 2016 From: India Posts: 14 Thanks: 0 Fractional Exponents Hi Experts Can you please help me to find Fractional exponents value For Eg: 2^1/3 Thank You in advance
 March 10th, 2016, 02:36 AM #2 Senior Member   Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics $\displaystyle 2^{1/3} = \sqrt[3]{2} \approx 1.26$
March 10th, 2016, 02:40 AM   #3
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Quote:
 Originally Posted by 123qwerty $\displaystyle 2^{1/3} = \sqrt[3]{2} \approx 1.26$
Value is okay sir but how you got it.what is procedural approach to solve such kind of problems

March 10th, 2016, 02:47 AM   #4
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Quote:
 Originally Posted by Prudhvi raj k Value is okay sir but how you got it.what is procedural approach to solve such kind of problems
You use a calculator.

But if you really feel like fiddling around, then (for this case) you need to find a number that can be cubed (multiplied by itself 3 times) which will equal 2.

EDIT: Remember to not post your questions in the New Users section! Welcome to MMF

Please do not post math questions here

Last edited by Joppy; March 10th, 2016 at 02:51 AM.

 March 10th, 2016, 03:18 AM #5 Newbie   Joined: Mar 2016 From: India Posts: 14 Thanks: 0 okay sir and Is there any possibility to solve with logarithms ?
March 10th, 2016, 03:23 AM   #6
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Quote:
 Originally Posted by Prudhvi raj k Value is okay sir but how you got it.what is procedural approach to solve such kind of problems
There are many ways you could use I guess...

1) Let f(x) = 2^(1/3) - x. f(x) is continuous and differentiable everywhere on R, and you know that, say, f(0) > 0 and f(2) < 0. By the intermediate value theorem, you know the root is on (0, 2), so you apply the bisection method or Newton's method.

2) Let f(x) = 2^x. Find an appropriate Taylor approximation, i.e. f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ... for x=1/3 and a = 0.

March 10th, 2016, 03:25 AM   #7
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Quote:
 Originally Posted by Prudhvi raj k okay sir and Is there any possibility to solve with logarithms ?
Doesn't that just make things even more complicated?

And actually, my Taylor approximation suggestion has lots of ln 2s in it, so I guess it uses logarithms

Last edited by 123qwerty; March 10th, 2016 at 03:27 AM.

 March 10th, 2016, 06:18 AM #8 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,417 Thanks: 1025 8^(1/3) = 2 : digging a hole where all is soft earth... 2^(1/3) = ~1.26 : digging hole where earth is crusty, and lots of stones

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