Question: What is the domain of the function $f(x) = \frac{2-x}{\log(2-\log(x-2))}$, where $\log$ is the base $10$ logarithm function? Express your answer in interval notation.

To determine the domain of the function \( f(x) = \frac{2-x}{\log(2-\log(x-2))} \), we need to ensure that both the numerator and the denominator are defined and that the denominator is not zero. Let's break this down step by step.

### Step 1: Ensure the argument of the inner logarithm is positive
The inner logarithm is \(\log(x-2)\). For the logarithm to be defined, its argument must be positive:
\[
x - 2 > 0 \implies x > 2
\]

### Step 2: Ensure the argument of the outer logarithm is positive
The outer logarithm is \(\log(2 - \log(x-2))\). For this logarithm to be defined, its argument must be positive:
\[
2 - \log(x-2) > 0 \implies \log(x-2) < 2
\]
Since \(\log(x-2) < 2\), we convert this inequality to exponential form:
\[
x - 2 < 10^2 \implies x - 2 < 100 \implies x < 102
\]

### Step 3: Combine the conditions from Steps 1 and 2
From Step 1, we have \(x > 2\). From Step 2, we have \(x < 102\). Therefore, combining these two conditions, we get:
\[
2 < x < 102
\]

### Step 4: Ensure the denominator is not zero
The denominator of the function is \(\log(2 - \log(x-2))\). For the function to be defined, the denominator must not be zero:
\[
\log(2 - \log(x-2)) \neq 0
\]
This implies:
\[
2 - \log(x-2) \neq 1 \implies \log(x-2) \neq 1
\]
Converting this inequality to exponential form:
\[
x - 2 \neq 10^1 \implies x - 2 \neq 10 \implies x \neq 12
\]

### Step 5: Combine all conditions
From Step 3, we have \(2 < x < 102\). From Step 4, we have \(x \neq 12\). Therefore, the domain of the function is:
\[
(2, 12) \cup (12, 102)
\]

Thus, the domain of the function \( f(x) = \frac{2-x}{\log(2-\log(x-2))} \) is \(\boxed{(2, 12) \cup (12, 102)}\).


Question: Let $z = 1+i$ and $w = \dfrac{3z+1}{5z+7}$. Find $|w|$.

To find the magnitude \( |w| \) where \( w = \frac{3z + 1}{5z + 7} \) and \( z = 1 + i \), we need to follow these steps:

1. **Substitute \( z = 1 + i \) into the expression for \( w \):**

   \[
   w = \frac{3(1 + i) + 1}{5(1 + i) + 7}
   \]

2. **Simplify the numerator and the denominator:**

   - Numerator: \( 3(1 + i) + 1 = 3 + 3i + 1 = 4 + 3i \)
   - Denominator: \( 5(1 + i) + 7 = 5 + 5i + 7 = 12 + 5i \)

   So,

   \[
   w = \frac{4 + 3i}{12 + 5i}
   \]

3. **To find the magnitude \( |w| \), use the property of magnitudes of complex numbers:**

   \[
   |w| = \left| \frac{4 + 3i}{12 + 5i} \right| = \frac{|4 + 3i|}{|12 + 5i|}
   \]

4. **Calculate the magnitudes of the numerator and the denominator:**

   - Magnitude of \( 4 + 3i \):

     \[
     |4 + 3i| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5
     \]

   - Magnitude of \( 12 + 5i \):

     \[
     |12 + 5i| = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13
     \]

5. **Divide the magnitudes:**

   \[
   |w| = \frac{5}{13}
   \]

Therefore, the magnitude \( |w| \) is \(\boxed{\frac{5}{13}}\).