The problem asks to simplify four expressions using only positive exponents and without a calculator. a. $(9^{1/2}x^2y)(27^{1/3}y^{-1})$ b. $(x^{1/2})^{-2}$ c. $(\frac{1}{125})^{2/3}$ d. $\frac{8x^3}{-2x^{-2}}$

1. Problem Description

The problem asks to simplify four expressions using only positive exponents and without a calculator.

a. $(9^{1/2}x^2y)(27^{1/3}y^{-1})$

b. $(x^{1/2})^{-2}$

c. $(\frac{1}{125})^{2/3}$

d. $\frac{8x^3}{-2x^{-2}}$

2. Solution Steps

a. $(9^{1/2}x^2y)(27^{1/3}y^{-1})$

First, evaluate $9^{1/2}$ and $27^{1/3}$:

$9^{1/2} = \sqrt{9} = 3$

$27^{1/3} = \sqrt[3]{27} = 3$

Substitute these values into the expression:

$(3x^2y)(3y^{-1})$

Multiply the terms:

$3 \cdot 3 \cdot x^2 \cdot y \cdot y^{-1}$

$9x^2y^{1+(-1)}$

$9x^2y^0$

Since $y^0 = 1$, we have:

$9x^2$

b. $(x^{1/2})^{-2}$

Using the power of a power rule, $(a^m)^n = a^{m \cdot n}$:

$x^{(1/2) \cdot (-2)}$

$x^{-1}$

To express with positive exponent:

$x^{-1} = \frac{1}{x}$

c. $(\frac{1}{125})^{2/3}$

$(\frac{1}{125})^{2/3} = \frac{1^{2/3}}{125^{2/3}}$

$1^{2/3} = 1$

$125^{2/3} = (125^{1/3})^2 = (\sqrt[3]{125})^2 = (5)^2 = 25$

Therefore,

$\frac{1}{25}$

d. $\frac{8x^3}{-2x^{-2}}$

Divide the coefficients:

$\frac{8}{-2} = -4$

Divide the variables:

$\frac{x^3}{x^{-2}} = x^{3 - (-2)} = x^{3+2} = x^5$

Combine:

$-4x^5$

Since we want only positive exponents, we write the simplified expression as:

$-4x^5$

3. Final Answer

a. $9x^2$

b. $\frac{1}{x}$

c. $\frac{1}{25}$

d. $-4x^5$