Question 1204990
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a = 116 since it's the coefficient out front.
Alternatively, plug in t = 0 to find f(0) = 116 in the first equation and f(0) = a in the second equation. That leads to a = 116.



{{{f(t) = 116*e^(kt)}}} rewrites to {{{f(t) = 116*(e^k)^t}}}
The base is {{{e^k}}}
Set this equal to the base 0.69 and isolate k.
{{{e^k = 0.69}}}
{{{k = ln(0.69)}}}
{{{k = -0.371}}} approximately



<font color=red>Summary</font>
a = 116
k = -0.317 approximately
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