Let `f(x)={{:((log(1+ax)-log(1-bx))/x, x ne 0), (k,x=0):}`.
Find 'k' so that f(x) is continuous at x = 0.

Answer

To find the value of \( k \) such that the function \( f(x) \) is continuous at \( x = 0 \), we need to ensure that:

\[ \lim_{x \to 0} f(x) = f(0) \]

Given the function:

\[ f(x) = \frac{\log(1 + ax) - \log(1 - bx)}{x} \quad \text{for } x \neq 0 \]

and

\[ f(0) = k \]

### Step 1: Calculate the limit as \( x \) approaches 0

We need to evaluate:

\[ \lim_{x \to 0} \frac{\log(1 + ax) - \log(1 - bx)}{x} \]

### Step 2: Apply L'Hôpital's Rule

Since substituting \( x = 0 \) directly gives us the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. This requires us to differentiate the numerator and the denominator.

**Numerator:**

Differentiate \( \log(1 + ax) - \log(1 - bx) \):

\[ \frac{d}{dx}[\log(1 + ax)] = \frac{a}{1 + ax} \]
\[ \frac{d}{dx}[\log(1 - bx)] = \frac{-b}{1 - bx} \]

Thus, the derivative of the numerator is:

\[ \frac{a}{1 + ax} + \frac{b}{1 - bx} \]

**Denominator:**

The derivative of \( x \) is \( 1 \).

### Step 3: Rewrite the limit

Now we can rewrite the limit as:

\[ \lim_{x \to 0} \left( \frac{a}{1 + ax} + \frac{b}{1 - bx} \right) \]

### Step 4: Substitute \( x = 0 \)

Now substitute \( x = 0 \) into the limit:

\[ \frac{a}{1 + a \cdot 0} + \frac{b}{1 - b \cdot 0} = \frac{a}{1} + \frac{b}{1} = a + b \]

### Step 5: Set the limit equal to \( k \)

For the function to be continuous at \( x = 0 \):

\[ k = \lim_{x \to 0} f(x) = a + b \]

### Final Result

Thus, the value of \( k \) is:

\[ \boxed{a + b} \]