Let the domain of the function $f(x) = \cos^{-1} \left( \frac{4x + 5}{3x - 7} \right)$ be $[\alpha, \beta]$ and the domain of $g(x) = \log_2(2 - 6\log_7(2x + 5))$ be $[\gamma, \delta]$. Then $|7(\alpha + \beta) + 4(\gamma + \delta)|$ is equal to _____

Step 1: Determine the domain of f(x)

For the inverse cosine function $\cos^{-1}(u)$ to be defined, its argument $u$ must satisfy $-1 \le u \le 1$. Therefore, we set up the inequality $-1 \le \frac{4x + 5}{3x - 7} \le 1$.

Step 2: Solve the inequality for f(x)

We solve the two inequalities separately. First, $\frac{4x + 5}{3x - 7} \ge -1$ simplifies to $\frac{7x - 2}{3x - 7} \ge 0$. The critical points are $x = \frac{2}{7}$ and $x = \frac{7}{3}$. Analyzing the sign changes, we get $x \in \left( -\infty, \frac{2}{7} \right] \cup \left( \frac{7}{3}, \infty \right)$. Second, $\frac{4x + 5}{3x - 7} \le 1$ simplifies to $\frac{x + 12}{3x - 7} \le 0$. The critical points are $x = -12$ and $x = \frac{7}{3}$. Analyzing the sign changes, we get $x \in \left[ -12, \frac{7}{3} \right)$. The intersection of these two solution sets gives the domain of $f(x)$ as $\left[ -12, \frac{2}{7} \right]$. Thus, $\alpha = -12$ and $\beta = \frac{2}{7}$.

Step 3: Determine the domain of g(x)

For the logarithm function $\log_b(u)$ to be defined, its argument $u$ must be positive. First, for $\log_2(2 - 6\log_7(2x + 5))$, we need $2 - 6\log_7(2x + 5) > 0$. This simplifies to $\log_7(2x + 5) < \frac{1}{3}$. Second, for $\log_7(2x + 5)$, we need $2x + 5 > 0$, which means $x > -\frac{5}{2}$. Combining $\log_7(2x + 5) < \frac{1}{3}$ with the base 7 (which is greater than 1), we get $2x + 5 < 7^{1/3}$, so $x < \frac{7^{1/3} - 5}{2}$. The intersection of $x > -\frac{5}{2}$ and $x < \frac{7^{1/3} - 5}{2}$ gives the domain of $g(x)$ as $\left( -\frac{5}{2}, \frac{7^{1/3} - 5}{2} \right)$. Thus, $\gamma = -\frac{5}{2}$ and $\delta = \frac{7^{1/3} - 5}{2}$.

Step 4: Calculate the final expression

Now we substitute the values of $\alpha, \beta, \gamma, \delta$ into the expression $|7(\alpha + \beta) + 4(\gamma + \delta)|$. First, calculate $\alpha + \beta = -12 + \frac{2}{7} = -\frac{82}{7}$. Next, calculate $\gamma + \delta = -\frac{5}{2} + \frac{7^{1/3} - 5}{2} = \frac{7^{1/3} - 10}{2}$. Substitute these sums into the given expression: $\left| 7\left(-\frac{82}{7}\right) + 4\left(\frac{7^{1/3} - 10}{2}\right) \right|$. Simplify the expression to get $| -82 + 2(7^{1/3} - 10) | = | -82 + 2 \cdot 7^{1/3} - 20 | = | -102 + 2 \cdot 7^{1/3} |$.

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