(12x^2*y^6)^2/8x^4*y^7

Question

Please solve!

 (12x²·y⁶)² 

      over       

   8x⁴·y⁷ 

(twelve 'x' to the second power times 'y' to the sixth power) to the power of two OVER eight 'x' to the fourth power times 'y' to the seventh power

 

Answer 1

In the numerator, square everything inside the parentheses:  12² = 144; then recall that when you raise a power to a power, you multiply.  So (x²)² = x⁴, and (y⁶)² = y¹².  So now you have 144x⁴y¹²  /  8x⁴y⁷

Now divide the numerator by the denominator:  144/8 (then remember that when you divide exponents, you subtract)  x⁴/x⁴  (4-4 = 0, or another way to look at it is that any number divided by the same number equals one) and any number raised to the power of zero = 1, and the x values cancel each other out; y¹²/y⁷  (12-7 = 5).  So final answer is

18y⁵