Question:

Step 1: Identify the domain. For $\log_2(x + 1)$ to be defined, the argument must be positive. $$x + 1 > 0 \implies x > -1$$ Step 2: Break the compound inequality into two parts. The given inequality is $5x - 1 < \log_2(x + 1) \le 2x - 3$. This can be separated into two distinct inequalities: (i) $5x - 1 < \log_2(x + 1)$ (ii) $\log_2(x + 1) \le 2x - 3$ Step 3: Solve inequality (i): $5x - 1 < \log_2(x + 1)$. Consider the function $f(x) = \log_2(x + 1) - (5x - 1)$. We need to find $x$ such that $f(x) > 0$. Let's evaluate $f(x)$ at some points within the domain $x > -1$: For $x = 0$: $f(0) = \log_2(1) - (5(0) - 1) = 0 - (-1) = 1$. Since $1 > 0$, $x=0$ satisfies the inequality. For $x = 1$: $f(1) = \log_2(2) - (5(1) - 1) = 1 - 4 = -3$. Since $-3 < 0$, $x=1$ does not satisfy the inequality. Since $f(x)$ is continuous for $x > -1$ and changes sign between $x=0$ and $x=1$, there exists a unique root $x_1 \in (0, 1)$ such that $f(x_1) = 0$. For $x \in (-1, x_1)$, $f(x) > 0$, meaning $5x - 1 < \log_2(x + 1)$. Thus, the solution for inequality (i) is $x \in (-1, x_1)$, where $x_1 \approx 0.265$. Step 4: Solve inequality (ii): $\log_2(x + 1) \le 2x - 3$. Consider the function $g(x) = 2x - 3 - \log_2(x + 1)$. We need to find $x$ such that $g(x) \ge 0$. Let's evaluate $g(x)$ at some points within the domain $x > -1$: For $x = 2$: $g(2) = 2(2) - 3 - \log_2(2 + 1) = 1 - \log_2(3) \approx 1 - 1.585 = -0.585$. Since $-0.585 < 0$, $x=2$ does not satisfy the inequality. For $x = 3$: $g(3) = 2(3) - 3 - \log_2(3 + 1) = 3 - \log_2(4) = 3 - 2 = 1$. Since $1 \ge 0$, $x=3$ satisfies the inequality. Since $g(x)$ is continuous for $x > -1$ and changes sign between $x=2$ and $x=3$, there exists a unique root $x_2 \in (2, 3)$ such that $g(x_2) = 0$. For $x \ge x_2$, $g(x) \ge 0$, meaning $\log_2(x + 1) \le 2x - 3$. Thus, the solution for inequality (ii) is $x \in [x_2, \infty)$, where $x_2 \approx 2.375$. Step 5: Find the intersection and count integers. The solution set for the compound inequality is the intersection of the solutions from Step 3 and Step 4, combined with the domain from Step 1. From Step 3, the solution is $x \in (-1, x_1)$, where $x_1 \in (0, 1)$. From Step 4, the solution is $x \in [x_2, \infty)$, where $x_2 \in (2, 3)$. Since $x_1 < 1$ and $x_2 > 2$, it is clear that $x_1 < x_2$. Therefore, the intersection of the two intervals $(-1, x_1)$ and $[x_2, \infty)$ is empty. There are no values of $x$ that satisfy the given compound inequality. The number of integral values of $x$ is $0$.