Question

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

Answer 1

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`