Find the value of $10^x$, given that $4^{x+1} \times 5^{2x-3} = 10^{3x}$. Hence, solve $4^{x+1} \times 5^{2x-3} = 10^{3x}$.

1. Express all terms using prime bases $2$ and $5$.

$(2^2)^{x+1} \times 5^{2x-3} = (2 \times 5)^{3x}$
$2^{2x+2} \times 5^{2x-3} = 2^{3x} \times 5^{3x}$

2. Group the terms with the same base on each side.

$\frac{2^{2x+2}}{2^{3x}} \times \frac{5^{2x-3}}{5^{3x}} = 1$
$2^{2x+2-3x} \times 5^{2x-3-3x} = 1$
$2^{2-x} \times 5^{-3-x} = 1$

3. Isolate $10^x$ to find its value.

$2^2 \cdot 2^{-x} \times 5^{-3} \cdot 5^{-x} = 1$
$\frac{4}{125} \times (2 \times 5)^{-x} = 1$
$\frac{4}{125} \times 10^{-x} = 1$
$10^{-x} = \frac{125}{4}$
$10^x = \frac{4}{125}$

1. Find the value of $10^x$ from part (i)

From part (i):
$4^{x+1} \times 5^{2x-3} = 10^{3x}$
$4(4^x) \times \frac{1}{125}(25^x) = 1000^x$
$\frac{4}{125}(100^x) = 1000^x$
$10^x = \frac{4}{125} = 0.032$

2. Solve for $x$ using logarithms

$10^x = 0.032$
$x = \lg(0.032)$
$x \approx -1.49485 \approx -1.49 \text{ (to 3 s.f.)}$