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Given the exponential function f(x) = e^kt, what is the value for k when t = 20 seconds and f(x) = 1.224

 Apr 16, 2014
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1.224 = ek*20 Take logs of both sides: ln(1.224) = k*20;  rearrange: k = ln(1.224)/20

$${\mathtt{k}} = {\frac{{ln}{\left({\mathtt{1.224}}\right)}}{{\mathtt{20}}}} = {\mathtt{k}} = {\mathtt{0.010\: \!106\: \!209\: \!204\: \!506\: \!7}}$$

 so k is approximately equal to 0.01 per second. 

 Apr 16, 2014

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