Question

Find the values of a, b and c in the equation below (x^5yz^4)^3/x^3yz= x^a y^b z^c

Answer 1

Answer: a = 12, b = 2, c = 11

Step-by-step explanation:

Using exponent rules we can distribute the exponent of 3 to the xyz.

This results in the numerator now being. $x^{5*3}+y^{1*3}+z^{4*3}$ This can be simplified to, $x^{15}y^{3}z^{12}$.

Now divide the numerator by the denominator subtracting the exponents.

Which equals, $x^{12}y^{2}z^{11}$

So a = 12

b = 2

and

c = 11