\( \Rightarrow \frac { e ^ { x } } { e ^ { x } + 1 } = \ln a \)

To solve the equation \[ \frac{e^x}{e^x + 1} = \ln a, \] we can start by isolating \( e^x \). First, let's multiply both sides by \( e^x + 1 \): \[ e^x = \ln a (e^x + 1). \] Expanding the right side gives: \[ e^x = \ln a \cdot e^x + \ln a. \] Next, we can rearrange the equation to isolate terms involving \( e^x \): \[ e^x - \ln a \cdot e^x = \ln a. \] Factoring out \( e^x \) from the left side results in: \[ e^x (1 - \ln a) = \ln a. \] Now, we can solve for \( e^x \): \[ e^x = \frac{\ln a}{1 - \ln a}. \] Taking the natural logarithm of both sides gives: \[ x = \ln\left(\frac{\ln a}{1 - \ln a}\right). \] Thus, the solution for \( x \) in terms of \( a \) is: \[ x = \ln\left(\frac{\ln a}{1 - \ln a}\right). \] This is valid as long as \( \ln a \neq 1 \) (which would make the denominator zero) and \( \ln a > 0 \) (which implies \( a > 1 \)).