Given that $4(5^{x+3}) = 20^{3-x}$, evaluate $10^x$ without using a calculator.

1. Express the base 20 in terms of its prime factors 4 and 5, and apply the power of a product rule.

$4(5^{x+3}) = (4 \times 5)^{3-x}$
$4^1 \cdot 5^{x+3} = 4^{3-x} \cdot 5^{3-x}$

2. Group terms with the same base on the same side of the equation.

$\frac{4^1}{4^{3-x}} = \frac{5^{3-x}}{5^{x+3}}$
$4^{1 - (3-x)} = 5^{(3-x) - (x+3)}$

3. Simplify the exponents and rearrange to form a common power.

$4^{x-2} = 5^{-2x}$
$4^x \cdot 4^{-2} = (5^{-2})^x$
$\frac{4^x}{16} = \left(\frac{1}{25}\right)^x$
$4^x \cdot 25^x = 16$

4. Combine the bases to form $100^x$ and solve for $10^x$.

$(4 \times 25)^x = 16$
$100^x = 16$
$(10^2)^x = 16$
$(10^x)^2 = 16$
Since $10^x > 0$ for all real $x$,
$10^x = \sqrt{16} = 4$