Question

last, simplify $4^{12} \cdot 4^{-2}$.
use the product of \underline{\hspace{2cm}} powers property.
$= 4^{12} \cdot 4^{-2}$
$= 4^{\underline{\hspace{0.5cm}} + \underline{\hspace{0.5cm}}}$
$= 4^{\underline{\hspace{0.5cm}}}$
$(4^3)^4 \cdot (4^{-2} \cdot 3^0) = \underline{\hspace{2cm}}$

Explanation:

Step1: Apply product of powers rule

For $4^{12} \cdot 4^{-2}$, use $a^m \cdot a^n = a^{m+n}$:
$4^{12 + (-2)}$

Step2: Simplify the exponent

Calculate the sum of exponents:
$4^{12 - 2} = 4^{10}$

Step3: Simplify the second expression

First, handle $(4^3)^4$ with $(a^m)^n = a^{m \cdot n}$:
$(4^3)^4 = 4^{3 \cdot 4} = 4^{12}$
Then, $3^0 = 1$, so $4^{-2} \cdot 3^0 = 4^{-2} \cdot 1 = 4^{-2}$

Step4: Combine using product rule

Multiply $4^{12} \cdot 4^{-2}$:
$4^{12 + (-2)} = 4^{10}$

Answer:

For $4^{12} \cdot 4^{-2}$: $4^{10}$
For $(4^3)^4 \cdot (4^{-2} \cdot 3^0)$: $4^{10}$