View Full Version : Would someone please help me with some basic algebra?
How would I simplify this equation? (3a^3)^-3(9a^-1)^-2
When I put those carrots in that means the next number is an exponent. Is there any easier way to write exponents?
How would I simplify this one: (2a^-1)^-2(2a^-1)^4
Would the simplified expression of the second equation be 4/a ?
Carrot :biggrin: That's "caret", btw :wink:
Anyway, "distribute" some of the exponents. For example, (2a^{-1})^{-2} = 2^{-2} \cdot a^{(-1)(-2)} = \frac{1}{2^2} \cdot a^2 = a^2/4 . Then do the same for the other paranthesis, multiply them together, etc.
The last one is almost 4/a, it's 4/a^2...
HallsofIvy
Jun29-04, 06:42 PM
As far as "(2a^-1)-2(2a^-1)4" is concerned, there is a general formula: axay= ax+y.
(Edit: replaced x*y with x+y!)
loseyourname
Jun29-04, 06:45 PM
When I put those carrots in that means the next number is an exponent. Is there any easier way to write exponents?
You can use LaTeX typesetting. Refer to this thread: http://www.physicsforums.com/showthread.php?t=8997
Also, you can click on the expressions that Muzza typed out to see how he did it.
hello3719
Jun29-04, 10:18 PM
As far as "(2a^-1)-2(2a^-1)4" is concerned, there is a general formula: axay= ax*y.
you mean axay= ax+y
ah nice hello. I thought that I was the only one that caught that. That would be sad. Thanks guys, this really helps, sorry if I bore you all with these simple problems, but i have to start somewhere.
As far as "(2a^-1)-2(2a^-1)4" is concerned, there is a general formula: axay= ax*y.
Do you perhaps mean (ax)y = ax*y?
HallsofIvy
Jun30-04, 06:03 AM
Yes, I just corrected it!
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