SOLUTION: i hope you can solve the following for me: 1. x2m+n (2m+n being the exponent) ------ xm+3n (m+3n being the exponent) 2. 22m+2 (2m+2 being the exponent)

Algebra ->  Exponents -> SOLUTION: i hope you can solve the following for me: 1. x2m+n (2m+n being the exponent) ------ xm+3n (m+3n being the exponent) 2. 22m+2 (2m+2 being the exponent)       Log On


   

Question 441213: i hope you can solve the following for me:
1. x2m+n (2m+n being the exponent)
------
xm+3n (m+3n being the exponent)

2. 22m+2 (2m+2 being the exponent)
-----
2(2m-1)2 ( (2m-1)2 being the exponent)

3. (x+1)3n-2 (3n-2 being the exponent)
---------
(x+1)2-3n (2-3n being the exponent)


THANK YOU
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi, Note the  use of ^ (uppercase 6)

Refer to the Laws governing exponents:

%28a%5Ep%29%5Eq+=+a%5E%28p%2Aq%29**

root%28a%2Cp%29%2Aroot%28a%2Cq%29+=+root%28a%2Cp%2Aq%29

x%5Em%2Fx%5En=x%5E%28m-n%29 ****

%28+x%5Em+%29%28+x%5En+%29+=+x%5E%28+m+%2B+n+%29



1. x%5E%282m%2Bn%29%2Fx%5E%28m%2B3n%29+=+x%5E%282m%2Bn-m-3n%29+=+x%5E%28m-2n%29

2. 

3.