The inverse of the function `y =(10^x-10^(-x))/(10^x +10^(-x))+1 ` is

Options

A. `y= log_(10) (x )/(2-x)`
B. ` y=1/2 log_(10) (x/(2-x))`
C. `y=1/2 log_(10) (x)/(1-x)`
D. None of these

Answer

To find the inverse of the function \( y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}} + 1 \), we can follow these steps:

### Step 1: Rewrite the function
Let \( y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}} + 1 \).

### Step 2: Isolate the fraction
Subtract 1 from both sides:
\[ y - 1 = \frac{10^x - 10^{-x}}{10^x + 10^{-x}} \]

### Step 3: Apply the cross-multiplication
Cross-multiply to eliminate the fraction:
\[ (y - 1)(10^x + 10^{-x}) = 10^x - 10^{-x} \]

### Step 4: Expand the left side
Expanding gives:
\[ (y - 1)10^x + (y - 1)10^{-x} = 10^x - 10^{-x} \]

### Step 5: Rearrange the equation
Rearranging terms, we have:
\[ (y - 1)10^x - 10^x = - (y - 1)10^{-x} - 10^{-x} \]
This simplifies to:
\[ (y - 2)10^x = -(y + 1)10^{-x} \]

### Step 6: Multiply both sides by \( 10^x \)
To eliminate \( 10^{-x} \), multiply both sides by \( 10^x \):
\[ (y - 2)(10^x)^2 = -(y + 1) \]

### Step 7: Rearrange to form a quadratic equation
This gives us:
\[ (y - 2)(10^x)^2 + (y + 1) = 0 \]

### Step 8: Solve for \( 10^x \)
Let \( z = 10^x \). The equation becomes:
\[ (y - 2)z^2 + (y + 1) = 0 \]
This is a quadratic equation in \( z \).

### Step 9: Use the quadratic formula
Using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = (y - 2) \), \( b = 0 \), and \( c = (y + 1) \).
\[ z = \frac{-0 \pm \sqrt{0^2 - 4(y - 2)(y + 1)}}{2(y - 2)} \]
This simplifies to:
\[ z = \frac{\sqrt{-4(y - 2)(y + 1)}}{2(y - 2)} = \frac{\sqrt{4(2 - y)(y + 1)}}{2(y - 2)} = \frac{\sqrt{(2 - y)(y + 1)}}{y - 2} \]

### Step 10: Take logarithm to solve for \( x \)
Since \( z = 10^x \), we have:
\[ 10^x = \frac{\sqrt{(2 - y)(y + 1)}}{y - 2} \]
Taking logarithm base 10:
\[ x = \log_{10}\left(\frac{\sqrt{(2 - y)(y + 1)}}{y - 2}\right) \]

### Step 11: Substitute \( y \) with \( x \) to find the inverse
To express the inverse function, replace \( y \) with \( x \):
\[ f^{-1}(x) = \log_{10}\left(\frac{\sqrt{(2 - x)(x + 1)}}{x - 2}\right) \]

### Final Result
Thus, the inverse of the function is:
\[ f^{-1}(x) = \log_{10}\left(\frac{\sqrt{(2 - x)(x + 1)}}{x - 2}\right) \]