45 million \(2^{4}\) = 45million 16 = 720million
Or you can use
\(y=Ae^{kt}\)
A is equal to 45 million
To find the rate, k, we can plug in:
\(90million=45million*e^{k*1}\)
\(2=e^{k}\)
\(k=ln(2)\)
\(y=4.5*10^{7}*e^{t *ln2}\)
Knowing that t = 4
\(y = 4.5*10^{7}*e^{ln(2)*4}\)
\(y = 720 million\)