Put this in the form $y = mx +c$

Question

I need this equation in the $y = mx + c$ form, where $k$ is the slope.

I might need to take natural logarithms, but I am not sure?

$${C} = \{{1 - e^{-kt}}\} {A}$$

Answer 1

Not sure, if the question is of the form $C=1-Ae^{-kt}$, in which case, taking the $\log$ gets down to,

\begin{eqnarray*} \log (1-C) &=& \log\left(A e^{-kt}\right) \\ &=& \log A -k t \end{eqnarray*}

That is, $\log A = k t + \log(1-C) $ and is of the form $y=m t +c$, where $y=\log A, m=k $ and $c=\log(1-C)$.