Consider the following equation. \[ \frac{12}{1+e^{-x}}=2 \] (a) Find the exact solution of the exponential equation in terms of logarithms. \( x= \) (b) Use a calculator to find an approximation to the solution, rounded to six decimal places. \( x=\square \)

To solve the equation \(\frac{12}{1+e^{-x}}=2\), let's start by isolating the exponential term. First, multiply both sides by \(1 + e^{-x}\): \[ 12 = 2(1 + e^{-x}) \] Expanding the right side gives: \[ 12 = 2 + 2e^{-x} \] Now, subtract \(2\) from both sides: \[ 10 = 2e^{-x} \] Next, divide both sides by \(2\): \[ 5 = e^{-x} \] Taking the natural logarithm of both sides leads to: \[ -x = \ln(5) \] Therefore, we can express \(x\) as: \[ x = -\ln(5) \] Now, for the approximation, using a calculator, we can evaluate \( -\ln(5) \): \( x \approx -1.609438 \) (to six decimal places). So the answers are: (a) \( x = -\ln(5) \) (b) \( x = -1.609439 \)