Using Positive Exponents, how do you solve the following problems?

enter image source here

1 Answer
Mar 14, 2018
  1. 2x^4y^3
  2. 625a^8b^16
  3. 16777216x^30y^24
  4. 1

Explanation:

  1. (12x^7y^11)/(6x^3 y^8)
    Since everything on the numerator and denominator is multiplication, you simply divide like terms. When x^3 is divided by x, that is the same as (x*x*x)/x. The x in the denominator cancels out one of the x's in the numerator, leaving x^2. This implies that the exponent 1 in the denominator was subtracted from the exponent 3 in the numerator, leaving an exponent of 2: x^2.
    By this process, x^7/x^3=x^4 and y^11/y^8=y^3
    For the coefficients, 12/6 =2.
    Combined, that is 2x^4y^3

  2. (5a^2b^4)^4
    (5a^2b^4)^4=(5a^2b^4)*(5a^2b^4)*(5a^2b^4)*(5a^2b^4)
    When you raise an exponent to an exponent, the exponents multiply. The coefficient in the equation is also raised the to exponent. Thus,
    (5a^2b^4)^4=5^4*a^(2*4)*b^(4*4)=625a^8b^16

  3. (4^2x^5y^4)^6=(16x^5y^4)^6=16777216x^30y^24

  4. 32^0
    Any number (except 0) raised the the power of 0 is 1.
    32^0=1