Number of integers satisfying the inequality `log_((x + 3)/(x - 3))4 lt 2 [log_(1/2)(x - 3)-log_(sqrt(2)/2)sqrt(x + 3)]` is greater than

Options

A. a. `6`
B. b. `5`
C. c. `4`
D. d.`3`

Explanation

To solve the inequality

\[ \log_{\frac{x + 3}{x - 3}} 4 < 2 \left[ \log_{\frac{1}{2}} (x - 3) - \log_{\frac{\sqrt{2}}{2}} \sqrt{x + 3} \right], \]

we will follow these steps:

### Step 1: Rewrite the logarithmic terms
Using the change of base formula and properties of logarithms, we can rewrite the left side:

\[ \log_{\frac{x + 3}{x - 3}} 4 = \frac{\log 4}{\log \left(\frac{x + 3}{x - 3}\right)}. \]

Since \( \log 4 = 2 \log 2 \), we have:

\[ \log_{\frac{x + 3}{x - 3}} 4 = \frac{2 \log 2}{\log \left(\frac{x + 3}{x - 3}\right)}. \]

### Step 2: Simplify the right side
For the right side, we can rewrite the logarithms:

\[ \log_{\frac{1}{2}} (x - 3) = -\log_2 (x - 3), \]
and
\[ \log_{\frac{\sqrt{2}}{2}} \sqrt{x + 3} = \log_{2^{1/2}} (x + 3) = 2 \log_2 (x + 3). \]

Thus, the right side becomes:

\[ 2 \left[-\log_2 (x - 3) - 2 \log_2 (x + 3)\right] = -2 \log_2 (x - 3) + 2 \log_2 (x + 3). \]

### Step 3: Set up the inequality
Now we can set up the inequality:

\[ \frac{2 \log 2}{\log \left(\frac{x + 3}{x - 3}\right)} < -2 \log_2 (x - 3) + 2 \log_2 (x + 3). \]

### Step 4: Combine the logarithmic expressions
We can rewrite the right side as:

\[ 2 \left[\log_2 (x + 3) - \log_2 (x - 3)\right] = 2 \log_2 \left(\frac{x + 3}{x - 3}\right). \]

Thus, the inequality simplifies to:

\[ \frac{2 \log 2}{\log \left(\frac{x + 3}{x - 3}\right)} < 2 \log_2 \left(\frac{x + 3}{x - 3}\right). \]

### Step 5: Divide both sides by 2
Dividing both sides by 2 gives:

\[ \frac{\log 2}{\log \left(\frac{x + 3}{x - 3}\right)} < \log_2 \left(\frac{x + 3}{x - 3}\right). \]

### Step 6: Exponentiate both sides
Exponentiating both sides leads to:

\[ \left(\frac{x + 3}{x - 3}\right)^{\log 2} < \frac{x + 3}{x - 3}. \]

### Step 7: Rearranging the inequality
This can be rearranged to:

\[ \left(\frac{x + 3}{x - 3}\right)^{\log 2 - 1} < 1. \]

Since \( \log 2 < 1 \), we can conclude that:

\[ \frac{x + 3}{x - 3} < 1. \]

### Step 8: Solve the inequality
This implies:

\[ x + 3 < x - 3 \implies 3 < -3, \]

which is not possible. Thus, we need to consider the conditions under which the logarithms are defined.

### Step 9: Determine the valid range for x
The logarithm is defined for \( x - 3 > 0 \) and \( x + 3 > 0 \), which gives:

\[ x > 3. \]

### Step 10: Find integer solutions
Since \( x > 3 \), the integers satisfying this condition are \( 4, 5, 6, \ldots \).

### Conclusion
Thus, the number of integers satisfying the inequality is infinite, as there is no upper limit on \( x \).