The inverse of the function ` f(x) = log_a ( x+ sqrt(x^2 +1)) (a ge 0 , a ne 1)` is

Options

A. `1/2 (a^x -a^(-x))`
B. not defined for all x
C. defined for xgt0
D. none of these

Answer

To find the inverse of the function \( f(x) = \log_a (x + \sqrt{x^2 + 1}) \) where \( a \geq 0 \) and \( a \neq 1 \), we will follow these steps:

### Step 1: Set the function equal to \( y \)
We start by letting \( y = f(x) \):
\[ y = \log_a (x + \sqrt{x^2 + 1}) \]

### Step 2: Rewrite the logarithmic equation in exponential form
To eliminate the logarithm, we rewrite the equation in its exponential form:
\[ x + \sqrt{x^2 + 1} = a^y \]

### Step 3: Isolate the square root
Next, we isolate the square root term:
\[ \sqrt{x^2 + 1} = a^y - x \]

### Step 4: Square both sides
Now, we square both sides to eliminate the square root:
\[ x^2 + 1 = (a^y - x)^2 \]

### Step 5: Expand the right-hand side
Expanding the right side gives:
\[ x^2 + 1 = a^{2y} - 2a^y x + x^2 \]

### Step 6: Simplify the equation
We can simplify this by canceling \( x^2 \) from both sides:
\[ 1 = a^{2y} - 2a^y x \]

### Step 7: Rearrange to solve for \( x \)
Rearranging gives:
\[ 2a^y x = a^{2y} - 1 \]
\[ x = \frac{a^{2y} - 1}{2a^y} \]

### Step 8: Express \( x \) in terms of \( y \)
Thus, we have expressed \( x \) in terms of \( y \):
\[ x = \frac{1}{2} \left( a^y - a^{-y} \right) \]

### Step 9: Write the inverse function
Since we want the inverse function \( f^{-1}(x) \), we replace \( y \) with \( x \):
\[ f^{-1}(x) = \frac{1}{2} \left( a^x - a^{-x} \right) \]

### Final Answer
The inverse of the function \( f(x) = \log_a (x + \sqrt{x^2 + 1}) \) is:
\[ f^{-1}(x) = \frac{1}{2} \left( a^x - a^{-x} \right) \]

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