Help please negetive exponents?

Question

Im going to give three scenerios, all pointing the same direction, please help me i have an exam tommorow ill do anythingg 

. The problem statement, all variables and given/known data

http://i51.tinypic.com/2zyfyaa.jpg

Question 3 l )



2. Relevant equations



3. The attempt at a solution

(6a^-2b^-3)^2
(__________) 1/6a^2 * 1/b^3 / 1/2a^2 * 1/b^1 
(2a^2b^-1 )

I multiplied the top half together to give me 1/ 36a^4b^2

Multiplied the bottom half to give me 1/4a^4b^1 , Then took the reciprical of the bottom half and move it up and multiplied it with the top half.

Giving me a final of 

:: 4a^4b^1
__________
36^a4b^5

However, when i fold this down it becomes 1/9 and the a's cancel and 4b is left over at the bottom.

The answer is a^8b^4 / 9 

I know its complicated to read, just write it on paper and ull understand. 

What did i do wrong?

OR

Or.... 

Look here. 

http://www.purplemath.com/modules/simpexpo2.htm

Go down to Simplify the following expression: (–3x–1y2)2 

Why is it 9y^4 / x^2 , and not 9x^2/y^4 , doesnt make sense? anyone?

OR

... f) (-3m^-3n^-1)^-3

Solving this one, I get -27m^9 / n^3

Why in the answer book is it m^9*m^3 /27 ? 

Anyone....?

Please xplain to me what im doing wrong.. i use the neg reciprical rule and get the right answer but the order is all wrong..

PeaceLoveMusic · Accepted Answer

honestly I can't tell you but, if you go on Khan Academy or maybe even youtube, I'm sure you can find something helpful. Hope this helps. :)

Anonymous · Answer

*****Query 1*****
http://i51.tinypic.com/2zyfyaa.jpg   ....Question 3 - part (l)

[(6 a^-2 b^-3) / (2 a^2 b^-1)]^-2
= {[6/2] [(a^-2)/(a^2)] [(b^-3)/(b^-1)]}^-2
= [(6/2) a^(-2-2) b^(-3+1)]^-2
= [3 a^-4 b^-2]^-2
= (3^-2) a^(-4 * -2) b^(-2 * -2)
= [1/(3^2)] [a^8 b^4]
= (1/9)(a^8 b^4)

*****Query 2***** 
"Relevant equations"  ???  I have no idea what's your query here.

*****Query 3*****
You need to learn using multiple brackets generously  -the way you type the expressions can be very misleading!
I assume this is the expression you meant:
[(6 a^-2 b^-3)^2/(2 a^2 b^-1)] * [(1/(6a^2)) * (1/b^3)] / [(1/(2a^2)) * (1/b^1)]     ...see??  -it is neatly typed!

(6 a^-2 b^-3)^2/(2 a^2 b^-1)
= [(6^2) a^(-2 * 2) b^(-3 * 2)] / (2 a^2 b^-1)
= [ 36 a^-4 b^-6] / (2 a^2 b^-1)
= (36/2) a^(-4-2) b^(-6+1)
= 18 a^-6 b^-5

[(1/(6a^2)) * (1/b^3)] / [(1/(2a^2)) * (1/b^1)]
= [(1/6) (a^-2) (b^-3)] / [(1/2) (a^-2) (b^-1)]
= (2/6) a^(-2+2) b^(-3+1)
= (1/3) a^0 b^-2
= (1/3) b^-2

Now,
[(6 a^-2 b^-3)^2/(2 a^2 b^-1)] * [(1/6a^2) * (1/b^3)] / [(1/2a^2) * (1/b^1)]
= [18 a^-6 b^-5] / [(1/3) b^-2]
= (18 * 3) a^-6 b^(-5+2)
= 54 a^-6 b^-3
= 54 / (a^6 b^3)

*****Query 4*****
(–3x^-1 y^2)^2
= [(–3) (x^-1) (y^2)]^2
= (-3)^2 (x^-1)^2 (y^2)^2
= (9) x^(-1 * 2) y^(2 * 2)
= (9) (x^-2) (y^4)
= (9) (1/x^2) (y^4)
= 9 y^4 / x^2

*****Query 5*****
(-3 m^-3 n^-1)^-3
= [(-3) (m^-3) (n^-1)]^-3
= (-3)^-3  (m^-3)^-3  (n^-1)^-3
= [(-3)^3]^-1  m^[(-3) * (-3)]  n^[(-1) * (-3)]
= [-27]^-1  m^9  n^3
= (-1/27) m^9 n^3
= -(m^9 n^3)/27

Katy Willy says "...please help me i have an exam tommorow ***I'll DO ANYTHINGG***"

-Anything???  hmmm.....
Good luck in your exams!

Help please negetive exponents?

2 Answers