If lot 7- log 2 + log 16 - 2 log3 - log `"" (7)/(45)= 1 + log `k, find the value of k.

Answer

To solve the equation \( \log 7 - \log 2 + \log 16 - 2 \log 3 - \log \left(\frac{7}{45}\right) = 1 + \log k \), we will use properties of logarithms step by step.

### Step-by-step Solution:

1. **Apply the properties of logarithms**:
- Use the property \( \log a - \log b = \log \left(\frac{a}{b}\right) \) and \( \log a + \log b = \log(ab) \).
- Rewrite \( -\log 2 \) and \( -\log \left(\frac{7}{45}\right) \).

\[ \log 7 - \log 2 = \log \left(\frac{7}{2}\right) \]

\[ -\log \left(\frac{7}{45}\right) = \log \left(\frac{45}{7}\right) \]

So, the equation becomes:

\[ \log \left(\frac{7}{2}\right) + \log 16 - 2 \log 3 + \log \left(\frac{45}{7}\right) = 1 + \log k \]

2. **Combine the logarithms**:
- Now combine the logarithms on the left side:

\[ \log \left(\frac{7}{2} \cdot 16 \cdot \frac{45}{7}\right) - 2 \log 3 = 1 + \log k \]

- The term \( -2 \log 3 \) can be rewritten as \( \log(3^{-2}) = \log \left(\frac{1}{9}\right) \).

Thus, we have:

\[ \log \left(\frac{7 \cdot 16 \cdot 45}{2 \cdot 9}\right) = 1 + \log k \]

3. **Simplify the expression**:
- Calculate the product inside the logarithm:

\[ \frac{7 \cdot 16 \cdot 45}{2 \cdot 9} = \frac{7 \cdot 16 \cdot 5}{1} = 7 \cdot 80 = 560 \]

So, the equation now looks like:

\[ \log(560) = 1 + \log k \]

4. **Convert 1 to logarithmic form**:
- Recall that \( 1 = \log(10) \), so we can rewrite the equation:

\[ \log(560) = \log(10) + \log k \]

5. **Combine the logarithms**:
- Using the property \( \log a + \log b = \log(ab) \):

\[ \log(560) = \log(10k) \]

6. **Equate the arguments**:
- Since the logarithms are equal, their arguments must also be equal:

\[ 560 = 10k \]

7. **Solve for \( k \)**:
- Divide both sides by 10:

\[ k = \frac{560}{10} = 56 \]

### Final Answer:
The value of \( k \) is \( 56 \).

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