The number of integral values of x which are not in the domain of log2((x-7)/(2x-3)^2), is

Steps to Solve

1. Identify the domain restrictions for the logarithm

For the logarithmic function $\log_2\left(\frac{x-7}{(2x-3)^2}\right)$ to be defined, the argument must be greater than zero. Thus, we need to establish when:

$$ \frac{x-7}{(2x-3)^2} > 0 $$

2. Analyze the numerator and denominator

The numerator $x-7$ must be greater than zero, so we solve the inequality:

$$ x - 7 > 0 \implies x > 7 $$

For the denominator $(2x-3)^2$, since a square is always non-negative and cannot equal zero, we must ensure:

$$ (2x-3)^2 \neq 0 $$

This leads us to the condition:

$$ 2x - 3 \neq 0 \implies x \neq \frac{3}{2} $$

3. Combine the conditions

From the analysis, we see that:

• For $x-7 > 0$, we need $x > 7$
• The denominator condition gives $x \neq \frac{3}{2}$ (this does not impact our primary condition since $7 > \frac{3}{2}$)

Therefore, the only significant restriction is $x > 7$.

4. Determine the intervals of non-valid integral values of x

Since we need integer values not in the domain, we consider integers that are less than or equal to 7. The integers to consider are:

$$ 7, 6, 5, 4, 3, 2, 1, 0, -1, \ldots $$

The relevant integral values are from negative integers up to and including 7.

5. Count the integral values

The integral values of $x$ that are not in the domain are:

$$ { \ldots, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7 } $$

Counting these values gives us:

• From $-\infty$ to 7: Count all integers from negative infinity up to and including 7.

6. Final count

The total count of integral values not in the domain is $8$ (from $-\infty$ to $6$), plus $7$.