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 11-04-2004, 04:08 AM #1 davee Member   Registered: Oct 2002 Location: Ayrshire, Scotland Distribution: Suse(home) RHEL (Work) Posts: 262 Rep: How do you calculate (manually) 2 to the power 3.5? Ok - brain getting rusty here. 2^2 = 2 x 2 = 4 2^3 = 2 x 2 x 2 = 8 2^4 = 2 x 2 x 2 x 2 = 16 etc. How do I calculate 2^3.5? Don't need to know for any reason; it just started bugging me...! Dave
 11-04-2004, 04:18 AM #2 vharishankar Senior Member   Registered: Dec 2003 Posts: 3,142 Blog Entries: 4 Rep: We were taught how to do this at school. 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.
 11-04-2004, 04:25 AM #3 davee Member   Registered: Oct 2002 Location: Ayrshire, Scotland Distribution: Suse(home) RHEL (Work) Posts: 262 Original Poster Rep: I'll maybe do a little research then; it seems like something that should be easy to work out... Dave
 11-04-2004, 05:22 AM #4 bakery2k Newbie   Registered: Oct 2003 Posts: 0 Rep: 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)
11-04-2004, 05:51 AM   #5
vharishankar
Senior Member

Registered: Dec 2003
Posts: 3,142
Blog Entries: 4

Rep:

Quote:
 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...

 11-05-2004, 01:25 AM #6 Mathiasdm LQ Newbie   Registered: Aug 2004 Distribution: Ubuntu Gutsy Gibbon Posts: 25 Rep: Or slightly different: 2^(7/2) = 2^(7*1/2) = (2^7)^(1/2) = sqrt (128) = sqrt (64*2) = 8 sqrt 2 Only minor difference ;-)
 11-05-2004, 02:00 AM #7 Poetics Senior Member   Registered: Jun 2003 Location: California Distribution: Slackware Posts: 1,178 Rep: Blast this finding threads too late to contribute anything extra Nicely done, gents
 11-05-2004, 02:10 AM #8 kersten78 Member   Registered: Nov 2003 Location: Minneapolis, MN Distribution: Slackware, Debian, Gentoo, openSuSE Posts: 357 Rep: 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.
11-05-2004, 02:24 AM   #9
SciYro
Senior Member

Registered: Oct 2003
Location: hopefully not here
Distribution: Gentoo
Posts: 2,038

Rep:
Quote:
 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)
... 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

11-05-2004, 02:37 AM   #10
nuka_t
Member

Registered: Jun 2004
Location: Kalifornia
Distribution: YOPER+KDE
Posts: 263

Rep:
Quote:
 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...
simple???

<---needs to take calculus.

 11-05-2004, 02:56 AM #11 vharishankar Senior Member   Registered: Dec 2003 Posts: 3,142 Blog Entries: 4 Rep: 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. x^{0.5) = sqrt{x}
 11-05-2004, 11:22 AM #12 kersten78 Member   Registered: Nov 2003 Location: Minneapolis, MN Distribution: Slackware, Debian, Gentoo, openSuSE Posts: 357 Rep: 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...