(i) Separate variables and integrate:

\(\int \frac{1}{x} \, dx = \int \frac{e^{-t}}{k + e^{-t}} \, dt\)

\(\ln x = -\ln(k + e^{-t}) + C\)

Given \(x = 10\) when \(t = 0\), substitute to find \(C\):

\(\ln 10 = -\ln(k + 1) + C\)

Thus, \(C = \ln 10 + \ln(k + 1)\).

Therefore, the solution is:

\(\ln x - \ln 10 = -\ln(k + e^{-t}) + \ln(k + 1)\)

(ii) Substitute \(x = 20\), \(t = 1\) into the solution:

\(\ln 20 - \ln 10 = -\ln(k + e^{-1}) + \ln(k + 1)\)

Simplify and solve for \(k\):

\(\ln 2 = \ln \left( \frac{k + 1}{k + \frac{1}{e}} \right)\)

\(2(k + \frac{1}{e}) = k + 1\)

\(2k + \frac{2}{e} = k + 1\)

\(k = 1 - \frac{2}{e}\)

(iii) As \(t \to \infty\), \(e^{-t} \to 0\):

\(\ln x = -\ln k + \ln(k + 1)\)

\(x = \frac{k + 1}{k}\)

Substitute \(k = 1 - \frac{2}{e}\):

\(x = \frac{1 - \frac{2}{e} + 1}{1 - \frac{2}{e}} = \frac{2 - \frac{2}{e}}{1 - \frac{2}{e}}\)

\(x \to 48\) as \(t \to \infty\).