Let `f(x) = ln(x-1)(x-3)` and `g(x) = ln(x-1) + ln(x-3)` then,

Options

A. `f(x) = g(x) AA x`
B. `f(x) = g(x) AA x in (3, infty)`
C. `f(x) = g(x), AA x in (-infty,1) cup (3, infty)`
D. `f(x) = g(x) AA x in R -(1,3)`

Answer

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) given by:

1. \( f(x) = \ln((x-1)(x-3)) \)
2. \( g(x) = \ln(x-1) + \ln(x-3) \)

### Step 1: Simplifying \( g(x) \)

Using the property of logarithms that states \( \ln(a) + \ln(b) = \ln(ab) \), we can simplify \( g(x) \):

\[ g(x) = \ln(x-1) + \ln(x-3) = \ln((x-1)(x-3)) \]

### Step 2: Comparing \( f(x) \) and \( g(x) \)

Now we can see that:

\[ f(x) = \ln((x-1)(x-3)) = g(x) \]

This means that \( f(x) \) and \( g(x) \) are equal for all \( x \) where both functions are defined.

### Step 3: Finding the domain of \( f(x) \) and \( g(x) \)

To determine where these functions are defined, we need to find the values of \( x \) for which the arguments of the logarithms are positive.

- For \( g(x) \):
- \( x - 1 > 0 \) implies \( x > 1 \)
- \( x - 3 > 0 \) implies \( x > 3 \)

Thus, \( g(x) \) is defined for \( x > 3 \).

- For \( f(x) \):
- The same conditions apply since \( f(x) \) involves the product \( (x-1)(x-3) \).

Thus, \( f(x) \) is also defined for \( x > 3 \).

### Step 4: Conclusion

Since both functions are equal and defined for \( x > 3 \), we conclude that:

\[ f(x) = g(x) \text{ for } x > 3 \]

### Final Answer

The correct option is that \( f(x) = g(x) \) when \( x > 3 \).

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