Solve The Inequality $log_{10}(\frac{x+3}{2x-3} - \frac{10}{2x-3})^2 - 2log_{10}(\frac{2x+7}{2}) \geq 0$ And Find The Number Of Integer Solutions For X.

In this comprehensive exploration, we delve into the intricacies of solving a logarithmic inequality and determining the number of integral solutions for the variable x. The specific inequality we will address is:

$log_{10}(\frac{x+3}{2x-3} - \frac{10}{2x-3})^2 - 2log_{10}(\frac{2x+7}{2}) \geq 0$

This problem combines logarithmic functions, algebraic manipulation, and the crucial concept of domain restrictions. Our goal is to systematically break down the problem, clarify each step, and provide a detailed explanation to ensure a clear understanding of the solution process. This involves simplifying the logarithmic expression, addressing the domain restrictions imposed by logarithms and rational functions, and identifying the range of x values that satisfy the inequality. Ultimately, we will pinpoint the number of integer values within this range, thus solving the problem.

Logarithmic inequalities form a crucial part of mathematical problem-solving, particularly in algebra and calculus. These inequalities involve logarithmic functions, which are the inverse operations of exponentiation. To effectively solve logarithmic inequalities, a solid grasp of logarithm properties and domain restrictions is necessary. The domain of a logarithmic function is especially critical since logarithms are only defined for positive arguments. This means that the expression inside the logarithm must be greater than zero. Additionally, the base of the logarithm must be a positive number not equal to 1.

In the context of the given problem, $log_{10}(...)$ indicates a base-10 logarithm, commonly known as the common logarithm. Understanding that the argument of the logarithm must be positive is the first step in solving the inequality. Furthermore, we must consider any additional constraints imposed by rational expressions or other functions within the logarithmic argument. For example, in our inequality, we have rational expressions within the logarithm, such as $(\frac{x+3}{2x-3} - \frac{10}{2x-3})$, which introduce the requirement that the denominator $(2x-3)$ cannot be zero. By carefully considering these restrictions, we ensure that our solutions are valid within the defined mathematical space.

To methodically solve the given logarithmic inequality,

$log_{10}(\frac{x+3}{2x-3} - \frac{10}{2x-3})^2 - 2log_{10}(\frac{2x+7}{2}) \geq 0$,

we will proceed through a series of steps, ensuring each is clearly explained and justified. This will provide a comprehensive and understandable solution.

Step 1: Simplify the Argument of the First Logarithm

The first step involves simplifying the expression inside the square in the first logarithm:

$\frac{x+3}{2x-3} - \frac{10}{2x-3} = \frac{x+3-10}{2x-3} = \frac{x-7}{2x-3}$

So the inequality becomes:

$log_{10}(\frac{x-7}{2x-3})^2 - 2log_{10}(\frac{2x+7}{2}) \geq 0$

This simplification combines the fractions, making the argument of the logarithm more manageable. By reducing the complexity of the expression, we set the stage for applying logarithmic properties and solving the inequality.

Step 2: Apply Logarithmic Properties

Next, we will use the logarithmic property $log_b(a^c) = c \cdot log_b(a)$ to simplify the first term and the property $n \cdot log_b(a) = log_b(a^n)$ to handle the coefficient in the second term. Applying these properties, we get:

$2log_{10}|\frac{x-7}{2x-3}| - log_{10}(\frac{2x+7}{2})^2 \geq 0$

Note that we introduce the absolute value because squaring the fraction $(\frac{x-7}{2x-3})$ results in a positive value, regardless of the sign of the fraction itself. This is a critical detail to ensure we don't lose any solutions. Now, further simplifying the squared term inside the second logarithm:

$2log_{10}|\frac{x-7}{2x-3}| - log_{10}(\frac{(2x+7)^2}{4}) \geq 0$

These logarithmic properties are fundamental for manipulating and simplifying logarithmic expressions, and their correct application is essential for solving the inequality.

Step 3: Combine Logarithmic Terms

To combine the logarithmic terms, we use the property $log_b(a) - log_b(c) = log_b(\frac{a}{c})$. Therefore, the inequality becomes:

$log_{10}(\frac{(\frac{x-7}{2x-3})^2}{\frac{(2x+7)^2}{4}}) \geq 0$

This step involves a careful application of the division rule for logarithms. By combining the terms, we move closer to isolating x and finding the solution set. Simplification of the fraction inside the logarithm is the next step:

$log_{10}(\frac{4(x-7)^2}{(2x-3)^2} \cdot \frac{1}{(2x+7)^2}) \geq 0$

Step 4: Remove the Logarithm

To remove the logarithm, we use the fact that if $log_{10}(A) \geq 0$, then $A \geq 10^0 = 1$. This step is valid because the base of the logarithm is 10, which is greater than 1, so the logarithmic function is increasing. Thus, the inequality becomes:

$\frac{4(x-7)^2}{(2x-3)^2(\frac{(2x+7)^2}{1})} \geq 1$

Simplifying the inequality, we get:

$\frac{4(x-7)^2}{(2x-3)^2(2x+7)^2} \geq 1$

This transformation is crucial because it converts the logarithmic inequality into a standard algebraic inequality, which we can solve using algebraic techniques.

Step 5: Simplify the Inequality

Now, let's simplify the inequality:

$4(x-7)^2 \geq (2x-3)^2(\frac{(2x+7)^2}{1})$

However, notice that both sides of the inequality involve squared terms. This suggests we should first consider the conditions under which the original logarithmic expression is defined before proceeding with algebraic manipulations that might obscure the domain restrictions. Therefore, we will temporarily set aside this algebraic approach and focus on finding the domain.

Step 6: Determine the Domain

The domain of the original inequality is restricted by two primary considerations:

1. The arguments of the logarithms must be positive.
2. The denominators of any fractions must not be zero.

For the first logarithm, we need $(\frac{x-7}{2x-3})^2 > 0$. Since anything squared is non-negative, we just need to ensure $\frac{x-7}{2x-3} \neq 0$, which means $x \neq 7$ and $2x - 3 \neq 0$, so $x \neq \frac{3}{2}$. Also, we must ensure $2x - 3 \neq 0$, which we already accounted for.

For the second logarithm, we need $\frac{(2x+7)}{2} > 0$, which simplifies to $2x + 7 > 0$, so $x > -\frac{7}{2}$.

Combining these restrictions, the domain is $x > -\frac{7}{2}$, $x \neq \frac{3}{2}$, and $x \neq 7$.

Step 7: Revisit and Solve the Inequality with Domain Restrictions

Going back to the simplified inequality:

$\frac{4(x-7)^2}{(2x-3)^2(2x+7)^2} \geq 1$

We realize that directly solving this inequality is complex. Instead, let’s go back to the inequality before removing the logarithm:

$log_{10}(\frac{4(x-7)^2}{(2x-3)^2(2x+7)^2}) \geq 0$

Since the logarithm is base 10, this inequality is equivalent to:

$\frac{4(x-7)^2}{(2x-3)^2(2x+7)^2} \geq 1$

Let's analyze this inequality within the domain $x > -\frac{7}{2}$, $x \neq \frac{3}{2}$, and $x \neq 7$.

This inequality can be rewritten as:

$4(x-7)^2 \geq (2x-3)^2(2x+7)^2$

Taking the square root is not straightforward due to the complexity. Instead, we recognize that the left side is always non-negative. The right side is also always non-negative. We need to find when the left side is greater than or equal to the right side.

However, a more insightful approach is to consider when the logarithmic argument is equal to 1, as that’s the boundary where the logarithm changes sign:

$\frac{4(x-7)^2}{(2x-3)^2(\frac{(2x+7)^2}{1})} = 1$

$4(x-7)^2 = (2x-3)^2(2x+7)^2$

This equation is quite complex to solve analytically. Instead, let's consider the original inequality and test integer values within the domain.

Step 8: Test Integer Values within the Domain

The domain is $x > -\frac{7}{2} = -3.5$, $x \neq \frac{3}{2} = 1.5$, and $x \neq 7$. So, we consider integer values greater than -3.5, excluding 1.5 and 7. We test x = -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 8, etc.

For the inequality:

$log_{10}(\frac{4(x-7)^2}{(2x-3)^2(2x+7)^2}) \geq 0$

which is equivalent to:

$\frac{4(x-7)^2}{(2x-3)^2(2x+7)^2} \geq 1$

Let $f(x) = \frac{4(x-7)^2}{(2x-3)^2(2x+7)^2}$. We need to find integer values for which $f(x) \geq 1$.

• For x = -3: $f(-3) = \frac{4(-10)^2}{(-9)^2(1)^2} = \frac{400}{81} \approx 4.94 > 1$
• For x = -2: $f(-2) = \frac{4(-9)^2}{(-7)^2(3)^2} = \frac{324}{441} \approx 0.735 < 1$
• For x = -1: $f(-1) = \frac{4(-8)^2}{(-5)^2(5)^2} = \frac{256}{625} \approx 0.41 < 1$
• For x = 0: $f(0) = \frac{4(-7)^2}{(-3)^2(7)^2} = \frac{196}{441} \approx 0.44 < 1$
• For x = 1: $f(1) = \frac{4(-6)^2}{(-1)^2(9)^2} = \frac{144}{81} \approx 1.78 > 1$
• For x = 2: $f(2) = \frac{4(-5)^2}{(1)^2(11)^2} = \frac{100}{121} \approx 0.83 < 1$
• For x = 3: $f(3) = \frac{4(-4)^2}{(3)^2(13)^2} = \frac{64}{1521} \approx 0.042 < 1$
• For x = 4: $f(4) = \frac{4(-3)^2}{(5)^2(15)^2} = \frac{36}{5625} \approx 0.0064 < 1$
• For x = 5: $f(5) = \frac{4(-2)^2}{(7)^2(17)^2} = \frac{16}{2021} \approx 0.0079 < 1$
• For x = 6: $f(6) = \frac{4(-1)^2}{(9)^2(19)^2} = \frac{4}{29161} \approx 0.00014 < 1$
• For x = 8: $f(8) = \frac{4(1)^2}{(13)^2(23)^2} = \frac{4}{84189} \approx 0.000047 < 1$

From the tested integer values, the inequality holds for x = -3 and x = 1.

Step 9: Finalize the Solution

Therefore, the integer solutions for x are -3 and 1. The number of integral solutions is 2.

The number of integral solutions for x is 2. Thus, the correct option is (A).

Solving logarithmic inequalities requires careful attention to the domain restrictions imposed by logarithmic functions and rational expressions. In this detailed solution, we systematically addressed the given inequality, simplifying the logarithmic expressions, identifying the domain, and testing integer values within the domain to find the solutions. The key steps included simplifying the arguments of the logarithms, applying logarithmic properties, determining the domain, and finally, testing integer values to identify the solutions. This comprehensive approach ensured that we accurately determined the number of integral solutions for x, highlighting the importance of both algebraic manipulation and domain considerations in solving logarithmic inequalities.