To solve the equation:

$\frac{1}{7^x} = 3 \times 49^{5 - x}$

and to express the answer in the form $a + \log_7 b$ where $a, b \in \mathbb{Z}$, let's proceed with the following steps.

1. Rewrite $\frac{1}{7^x}$ as $7^{-x}$, so the equation becomes: $7^{-x} = 3 \times 49^{5 - x}$

2. Express $49$ as $7^2$ so that we have consistent bases: $7^{-x} = 3 \times (7^2)^{5 - x}$

3. Simplify $(7^2)^{5 - x}$ to $7^{2(5 - x)} = 7^{10 - 2x}$. Substitute this into the equation: $7^{-x} = 3 \times 7^{10 - 2x}$

4. Now, equate the powers of 7 on both sides. Rewrite the equation as: $7^{-x} = 7^{10 - 2x} \times 3$

Since we need the form in