Show that the solution of the equation $12^{x+1} = 36 \times 4^{x+2}$ can be expressed as $1 + \frac{\lg a}{\lg b}$, where $a$ and $b$ are integers.

1. Express the equation using prime bases or common bases to simplify

$12^{x+1} = 36 \times 4^{x+2}$
$(3 \times 4)^{x+1} = 9 \times 4 \times 4^{x+2}$

2. Apply laws of indices to expand the terms

$3^{x+1} \times 4^{x+1} = 9 \times 4^{x+3}$

3. Group terms with the same base on opposite sides

$\frac{3^{x+1}}{9} = \frac{4^{x+3}}{4^{x+1}}$
$\frac{3^{x+1}}{3^2} = 4^{(x+3)-(x+1)}$
$3^{x-1} = 4^2$
$3^{x-1} = 16$

4. Take logarithms on both sides to solve for x

$\lg(3^{x-1}) = \lg 16$
$(x-1)\lg 3 = \lg 16$
$x - 1 = \frac{\lg 16}{\lg 3}$
$x = 1 + \frac{\lg 16}{\lg 3}$

5. Alternative grouping to match the exact form if needed (though the previous step gives a valid a and b)

Alternatively, from $3^{x+1} \times 4^{x+1} = 36 \times 4^{x+2}$:
$\frac{12^{x+1}}{4^{x+1}} = 36 \times 4$
$3^{x+1} = 144$
$(x+1)\lg 3 = \lg 144$
$x = -1 + \frac{\lg 144}{\lg 3}$ (not in the form $1 + \dots$)
Let's re-evaluate $3^{x-1} = 16 \implies x = 1 + \frac{\lg 16}{\lg 3}$. Here $a=16, b=3$, which are integers.