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Mathematics 13 Online
OpenStudy (lacris):

x^6a x^2/ x^2a please help me, I will give you a shiny medal! I keep getting different answers

jimthompson5910 (jim_thompson5910):

This is the problem given to you? \[\Large \frac{x^6*a*x^2}{x^2*a}\]

jimthompson5910 (jim_thompson5910):

Or is it this? \[\Large \frac{x^{6a}*x^2}{x^{2a}}\]

OpenStudy (lacris):

the second one :D

jimthompson5910 (jim_thompson5910):

ok the first thing we do is simplify the stuff up top

jimthompson5910 (jim_thompson5910):

focus on just \[\Large x^{6a}*x^2\] for now

jimthompson5910 (jim_thompson5910):

when we multiply exponential expressions with the same base, we add the exponents, so \[\Large x^{6a}*x^2 = x^{6a+2}\]

jimthompson5910 (jim_thompson5910):

making \[\Large \frac{x^{6a}*x^2}{x^{2a}}\] turn into \[\Large \frac{x^{6a+2}}{x^{2a}}\]

OpenStudy (lacris):

ooh, I thought we did 6a -2a first and then add 2 . so its 6a+2 - 2a?

jimthompson5910 (jim_thompson5910):

You could do that. I find it easier to simplify the numerator first. There are two ways to do this problem

OpenStudy (lacris):

so its x^6a+2/x^2a?

jimthompson5910 (jim_thompson5910):

Now subtract exponents \[\Large \frac{x^{6a+2}}{x^{2a}}=x^{(6a+2)-(2a)}\] \[\Large \frac{x^{6a+2}}{x^{2a}}=x^{6a+2-2a}\] \[\Large \frac{x^{6a+2}}{x^{2a}}=x^{6a-2a+2}\] \[\Large \frac{x^{6a+2}}{x^{2a}}=x^{4a+2}\]

jimthompson5910 (jim_thompson5910):

btw, when you want to say something like \(\Large x^{6a+2}\) make sure to use parenthesis and write x^(6a+2) saying x^6a+2 is a bit vague and it's hard to tell which is in the exponent or not

OpenStudy (lacris):

ahh I see now!, so we subtract from below since it's a division problem which is: division -> subtraction

OpenStudy (lacris):

And I didn't know about the parenthesis rule, thank you so much! :D

jimthompson5910 (jim_thompson5910):

yes, because of these exponent rules http://www.onlinemathlearning.com/image-files/xari121.gif.pagespeed.ic.vFV2Lan_oa.png rule 1 and rule 2

jimthompson5910 (jim_thompson5910):

you're welcome

OpenStudy (lacris):

thanks again :D

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