Let f be a real valued function, satisfying `f (x+y) =f (x) f (y)` for all a,y ` in R` Such that,` f (1_ =a. Then , f (x) = `

Options

A. `a ^(x)`
B. `ax`
C. `x ^(a)`
D. `log x`

Answer

To solve the problem, we need to find the function \( f(x) \) that satisfies the functional equation:

\[ f(x+y) = f(x)f(y) \]

for all \( x, y \in \mathbb{R} \), given that \( f(1) = a \).

### Step 1: Substitute \( y = 0 \)

Let's start by substituting \( y = 0 \) into the functional equation:

\[ f(x + 0) = f(x)f(0) \]

This simplifies to:

\[ f(x) = f(x)f(0) \]

### Step 2: Analyze the equation

We can rearrange the equation:

\[ f(x) - f(x)f(0) = 0 \]

Factoring out \( f(x) \):

\[ f(x)(1 - f(0)) = 0 \]

This implies either \( f(x) = 0 \) for all \( x \) (which is not a valid solution since \( f(1) = a \neq 0 \)), or \( 1 - f(0) = 0 \). Therefore, we have:

\[ f(0) = 1 \]

### Step 3: Substitute \( y = 1 \)

Next, we substitute \( y = 1 \) into the original equation:

\[ f(x + 1) = f(x)f(1) \]

Since \( f(1) = a \), we can rewrite this as:

\[ f(x + 1) = f(x)a \]

### Step 4: Find a pattern

Now, let's find \( f(2) \):

\[ f(2) = f(1 + 1) = f(1)f(1) = a \cdot a = a^2 \]

Next, find \( f(3) \):

\[ f(3) = f(2 + 1) = f(2)f(1) = a^2 \cdot a = a^3 \]

Continuing this pattern, we can see that:

\[ f(n) = a^n \quad \text{for any integer } n \]

### Step 5: Generalize for all \( x \)

To generalize this for all real \( x \), we can use the property of the function. The functional equation suggests that \( f(x) \) could be of the form \( a^x \).

### Step 6: Verify the solution

To verify, we check if \( f(x) = a^x \) satisfies the original equation:

\[ f(x+y) = a^{x+y} = a^x a^y = f(x)f(y) \]

This confirms that our solution is valid.

### Conclusion

Thus, the function \( f(x) \) is given by:

\[ f(x) = a^x \]