B raise to m times B raise to m...

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- Jun 18th 2006, 08:39 AM #1

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- Jun 2006
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- 2

- Jun 18th 2006, 08:46 AM #2Originally Posted by
**uzagie**

Chech with numbers.

$\displaystyle 3^2\cdot3^2=3^{(2+2)}=3^4=81 $

rules of powers:

$\displaystyle a^b\cdot a^c=a^{(b+c)} $

$\displaystyle \left(a^b\right)^c=a^{(b\cdot c)} $

$\displaystyle a^b+ a^b=2a^b $

$\displaystyle \left(a\cdot b\right)^c=a^c\cdot b^c $

and

$\displaystyle \left(a+b\right)^c$ can't be easily simplified

- Jun 18th 2006, 11:07 AM #3

- Joined
- Nov 2005
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- Wethersfield, CT
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- 92

Hi:

Of course you know that m+m=2m, and therefore*b^(m+m)*simplifies to b^(2m). Not a big deal but, perhaps enough save you a couple of points on the test/exam.

By the way, why is this question posted in two different locations within this site? As a result, three people have invested valuable time doing the job of one.

With regard to your posts, you may wish to recheck them for accuracy. They are not the same question, although I suspect they were intended to be.

Finally, the use of ellipses (...) when communicating mathematics should be approached with care. They carry specific meaning within the math world, and can therefore be misinturpreted by the reader. For instance, in the present case,

"*b raised to m times b raised to m...*" would indicate the expression goes on ad infinitum, giving

(b^m)(b^m)(b^m)(b^m)...= b^(m+m+m+...) = lim(b^(km))_[k->inf].

...so there you have it.

Regards,

Rich B.