How do I simplify fractions with exponents?

AI Thread Summary
To simplify the fraction (2x + 3x)/6x, start by splitting it into two separate fractions: (2x/6x) + (3x/6x). This can be simplified to 1/3 + 1/2, which can then be expressed as (1/2)^x + (1/3)^x by flipping the fractions and applying negative exponents. The key is recognizing that combining like terms and manipulating the fractions leads to a clearer form. Understanding how to apply exponent rules is essential in this process. Simplifying fractions with exponents can be straightforward with the right approach.
Aerospace93
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Homework Statement


(2x+3x)/6x = 2-x+3-x

I've tried moving the 6 above, splitting it up and so... but i can't figure how to do it. It must be pretty simple, but I am just not seeing it. all helps appreciated!
 
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Aerospace93 said:

Homework Statement


(2x+3x)/6x = 2-x+3-x

I've tried moving the 6 above, splitting it up and so... but i can't figure how to do it. It must be pretty simple, but I am just not seeing it. all helps appreciated!

Start by "splitting" the left side into two fractions, like this:
\frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}
 
i certainly did that
 
(1/2)^x+(1/3)^x... which is the same as flipping them over and making the exponential -x. cool
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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