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 hhmathgeek January 20th, 2016 05:07 PM

Exponent help

Hi :) I have this one math problem and it's really hard. I got an answer but it's probably wrong. Can someone please solve it and tell me what you get. I want to compare it to my answer :) Thanks. Here it is:

(2(squared) a(to the -1 power) b(to the -3 power) c) the -2 power

divided by 8( to the -1 power) ab( to the -5 power) c(to the 0 power)

*In the first line, all the terms from 2 to c are in parentheses and the -2 exponent is on the outside of the parentheses

 greg1313 January 20th, 2016 05:34 PM

Do you mean

$\displaystyle \dfrac{\left(2^2\cdot a^{-1}\cdot b^{-3}\cdot c\right)^{-2}}{8^{-1}\cdot(a\cdot b)^{-5} \cdot c^0}$

?

 hhmathgeek January 20th, 2016 05:38 PM

No, sorry, it's a little different; how did you type the exponents?

The exponent the a and b below the fraction bar are not in the same parentheses but other than that, yes :) that is what I meant.

Please solve :) if you get a chance. I really appreciate your help. Thanks.

 greg1313 January 20th, 2016 05:44 PM

I used LaTeX. Typing $2^{2}$ gives $\displaystyle 2^2$.

Typing $\dfrac{4}{3}$ gives $\displaystyle \dfrac{4}{3}$.

Typing $\dfrac{4}{3}+2^{2}$ gives $\displaystyle \dfrac{4}{3}+2^{2}$.

 greg1313 January 20th, 2016 05:51 PM

$\displaystyle \dfrac{\left(2^2\cdot a^{-1}\cdot b^{-3}\right)^{-2}}{8^{-1}\cdot a\cdot b^{-5}\cdot c^0}$

Like that?

I get $\displaystyle \dfrac12ab^{11}$.

 hhmathgeek January 20th, 2016 05:56 PM

Thank you so much. The only thing was you forgot the c in the parentheses above the fraction bar but that's okay. Thank you so much. I really appreciate it :)

 skipjack January 20th, 2016 05:57 PM

Quote:
 Originally Posted by hhmathgeek (Post 513969) I got an answer . . .

 greg1313 January 20th, 2016 06:00 PM

Quote:
 Originally Posted by hhmathgeek (Post 513988) The only thing was you forgot the c in the parentheses above the fraction bar...
Yep. :o

$\displaystyle \dfrac{\left(2^2\cdot a^{-1}\cdot b^{-3}\cdot c\right)^{-2}}{8^{-1}\cdot a\cdot b^{-5}\cdot c^0}=\dfrac12ab^{11}c^{-2}=\dfrac{ab^{11}}{2c^2}$

 hhmathgeek January 20th, 2016 06:03 PM

Thank you so much. My answer was sort of similar but not quite :)

 greg1313 January 20th, 2016 06:05 PM