
#1
Apr1806, 06:06 PM

P: 39

I was never taught how to do this.
How do i solve this. (x + 5)3 ? I'm thinking it's 15x? And for fractional exponenets. (27)2/3 =? and how do i solve for it W/O parenthesis 272/3 =? And if i get a mixed fractions like ? 



#2
Apr1806, 08:41 PM

P: 20

for your first problem, as it stands
(x + 5)3 = 3x + 15 but i think you meant to say [tex](x+5)^3[/tex] you can do this by applying the following twice [tex](a+b)^2=a^2+2ab+b^2[/tex] [tex](a+b)^3=a^3+3a^2b+3ab^2+b^3[/tex] so [tex](x+5)^3=x^3+15x^2+75x+125[/tex] 



#3
Apr1906, 03:46 AM

P: 1,157

You first problem is a bit strange as it just yields solutions at x=5
Your second question can be broken down a bit, basically by employing [tex]x^{ab}=(x^a)^b[/tex] and [tex]x^{a}=1/x^a[/tex] i.e. [tex](27)^{2/3}=(1)^{2/3}(27)^{2/3}=(1)^{2/3}((27)^{2})^{1/3}=(1)^{2/3}((1/27)^2)^{1/3}=(1)^{2/3}(1/729)^{1/3}=(1)^{2/3}(1/9)[/tex] The [tex](1)^{2/3}[/tex] bit requires you to know a bit about complex numbers too... 



#4
Apr1906, 02:24 PM

P: 81

Powers of sums/Fractional exponents
J77, why complicate the simple?
The second question doesn't require any knowledge about complex numbers, since (1)^{2/3} has an odd number in the denominator of the exponent. My experience is that people prefer to work with smaller numbers whenever possible. But maybe that's just me projecting my own preference onto others... (1/27)^{2/3} is a little easier to work with if you take the cube root before squaring  (1/3)^{2} = 1/9. Easier to take the cube root of 27 than the cube root of 729. Also easier to square 3 than to square 27. About your third question, Richay, a negative exponent means that number "wants" to be on the other side of the fraction bar. E.g. 2^{3} = 1/2^{3}. 1/5^{2} = 5^{2}. If the exponents are negative fractions, the same rule applies. You just have one extra step you need to take when evaluating the expression. 



#5
Apr1906, 02:27 PM

Sci Advisor
HW Helper
PF Gold
P: 12,016

Nimz:
And, what, may I ask, is [tex](1)^{\frac{4}{6}}[/tex] equal to? Do you still think that no knowledge of complex numbers is needed? 



#6
Apr2006, 03:18 AM

P: 1,157

Exactly, last time I looked...
[tex]1=e^{i\pi}[/tex] ie. [tex](1)^{2/3}=e^{2i\pi/3}=\cos(2\pi/3)i\sin(2\pi/3)\approx0.5i0.866[/tex] 



#7
Apr2006, 05:14 AM

Sci Advisor
HW Helper
P: 11,866

Actually
[tex] 1=e^{i\left(2k+1\right)\pi} , \forall{k\in\mathbb{Z}}[/tex] Daniel. 



#8
Apr2006, 04:18 PM

P: 81

My bad. I may have assumed too much in making the simplification of selecting k=1 (mod 3) in the formula 1=e^{i(2k+1)pi}. Incidentally, that is the same assumption made when making y=x^{1/(2n+1)} a function over the domain of all reals.
As to what (1)^{4/6} equals, it is precisely (1)^{2/3} (which can be any one of the three cube roots of unity). Back to the OP's questions, in general, (a)^{b} isn't the same as a^{b}, even with whole exponents. The former expression indicates multiplying (a) by itself b times, while the latter expression indicates multiplying a by itself b times, with a coefficient of 1. 


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