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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?

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

amWhy
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Eleftheria
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    There is no $x$ or $y$ in your formula, so how can we express it in the form you wish. Moreover, if $t$ is the variable of interest (is it?), then your equation is not linear and cannot be placed in the form you seek. – David G. Stork Sep 21 '18 at 17:01
  • Please use MathJax to format your math. The top-voted question on the meta site has a great tutorial. – John Sep 21 '18 at 17:03
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    Maybe you want $x=t$ but what is $y$? – user Sep 21 '18 at 17:23

1 Answers1

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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 $c=\log(1-C)$.

Nivedita Rethnakar
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