Questions: Apply all relevant Properties of Exponents to simplify the following expression so each base occurs only once and the exponents are simplified, nonnegative integers. a) (5 y)^3(6 y)^2 b) (1/9 q^3)^2(1/8 q^4)^2

Solution Steps

To simplify the given expressions using the properties of exponents, we will follow these steps:

1. Apply the power of a product property: \((ab)^n = a^n b^n\).
2. Apply the power of a power property: \((a^m)^n = a^{mn}\).
3. Combine like terms by adding or multiplying exponents as needed.

1. Distribute the exponents inside the parentheses.
2. Combine like terms by adding the exponents of \(y\).

1. Distribute the exponents inside the parentheses.
2. Combine like terms by adding the exponents of \(q\).
3. Simplify the coefficients by multiplying them.

We start with the expression \((5y)^3(6y)^2\). Applying the power of a product property, we have:

\[ (5y)^3 = 5^3y^3 = 125y^3 \] \[ (6y)^2 = 6^2y^2 = 36y^2 \]

Now, we can combine these results:

\[ 125y^3 \cdot 36y^2 = (125 \cdot 36)(y^3 \cdot y^2) = 4500y^{3+2} = 4500y^5 \]

Thus, the simplified expression for part (a) is:

\[ \boxed{4500y^5} \]

Next, we simplify the expression \(\left(\frac{1}{9}q^3\right)^2\left(\frac{1}{8}q^4\right)^2\). First, we apply the power of a product property:

\[ \left(\frac{1}{9}q^3\right)^2 = \frac{1^2}{9^2}q^{3 \cdot 2} = \frac{1}{81}q^6 \] \[ \left(\frac{1}{8}q^4\right)^2 = \frac{1^2}{8^2}q^{4 \cdot 2} = \frac{1}{64}q^8 \]

Now, we combine these results:

\[ \frac{1}{81}q^6 \cdot \frac{1}{64}q^8 = \left(\frac{1}{81} \cdot \frac{1}{64}\right)(q^{6+8}) = \frac{1}{5184}q^{14} \]

Calculating \(\frac{1}{81 \cdot 64}\):

\[ 81 \cdot 64 = 5184 \]

Thus, the simplified expression for part (b) is:

\[ \boxed{0.0001929q^{14}} \]

Final Answer