How do I solve this properties of exponets problem

Question

(-7)^2y^-3z^4/xy^2z^3

Putting it all together

(-7x)^2 y^-3 z^4

/xy^2z^3

*Its properties of exponents

Answer 1

Hi Eva,

This question is testing your knowledge of the division of exponents. Remember that when dividing exponents, you can just subtract the numerator's exponent by the denominator's exponent if the base number or variable is the same (e.g. both are x's). For example, how would I simplify:

x⁶ / x⁴

I would just subtract the numerator's exponent from the denominator's exponent:

x⁶ / x⁴ = x^(6-4) = x²

So for this question, I would do this for all the variables (x, y, z). So I would start by squaring -7, which is 49, and then subtracting the exponents for each variable. So the answer would be:

(-7x)² y^-3 z⁴ / x y² z³ = 49 x^(2-1) y^(-3-2) z^(4-3) = 49 x y^-5 z

And if you need to put your answer into all positive exponents, remember that to make an exponent positive, you just flip it to the opposite side of the fraction, so:

49 x y^-5 z = 49 x z / y⁵

I hope this helped!

Answer 2

The prompt just wants us to simplify. Remember that x^1 = x.

We have (-7x)^2 * y^(-3) * z^4 / ( x^1 * y^2 * z^3).

Part 1:

The first thing to do realize that (a*b)^2 = a^2 * b^2. So, (-7x)^2 = (-7)^2 * x^2 = 49 *x^2.

So, our expression is equivalent to 49 * x^2 * y^(-3) * z^4 / ( x^1 * y^2 * z^3).

Part 2:

Since division is, by definition, not an operation and is super painful to work with, we can convert the expression to only use multiplication. Remember that 1/x = x^(-1) and 1/(x^2) = x^(-2).

Thus, 49 * x^2 * y^(-3) * z^4 / ( x^1 * y^2 * z^3) = 49 * x^2 * y^(-3) * z^4 * (x^(-1) * y^(-2) * z^(-3)).

Now, we want to use the property that x^a * x^b = x^(a+b), along with commutativity.

We have then, 49 * x^(2+-1) * y^(-3+-2) * z^(4+-3) = 49 * x^(1) * y^(-5) * z^(1) = 49*x*z*y^(-5).

You could also rewrite this as 49xz/y^5.