How do you calculate (manually) 2 to the power 3.5?
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Apparently the only way to calculate the above is to use calculus. I forget the exact method, but somebody more in touch with Mathematics might be able to help you.
Originally posted by bakery2k 2^(3.5) = 2^(7*0.5) = (2^7)^(0.5) = 128^(0.5) = sqrt(128) = sqrt(64*2) = sqrt(64)*sqrt(2)
= 8 * sqrt(2)
Perfect mate! You got it. I was just thinking about it and was about to post this simple solution when I saw that somebody had beaten me to it. Thanks...
Much easier by addition of exponents. Since (a^x)*(a^y) = a^(x+y),
2^(3.5)=(2^3)*(2^.5), or 8*Sqrt[2]. Same concept as above, just using addition rule instead of multiplication rule. Sorry, I had to post, I'm a math nut.
... umm, since i never look calculus I'm a bit behind here .. but what happened to the "0.5" .. suddenly it vanished into 128 without doing a thing... i say the answer seems more like 12 if you count half's
Originally posted by Harishankar Perfect mate! You got it. I was just thinking about it and was about to post this simple solution when I saw that somebody had beaten me to it. Thanks...
The above is not the calculus method, everybody. It's simple math The calculus method is much tougher and is used by students specifically studying calculus. Not used by people who just want the answer.
Explanation
2^{3.5} = 2^{7 * 0.5} = (2^7)^{0.5) = 128^{0.5} (because 2 x 2 x 2 x 2 x 2 x 2 x 2 = 128)
Those who want the answer quickly usually use a calculator, not calculus
Sqrt is another way of saying square root which is same as raised to the power half.
What would the caculus method be? It seems like a basic algebra question regarding exponent rules... unless you want to find an approximation. Then maybe you could set up the continued fractions to approximate Sqrt[2], which still seems like an algebra problem, unless you consider some sort of limiting situation...which isn't really going to help since Sqrt[2] is irrational.
Hmm...maybe I missed the calculus of exponents. Who knows...
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