Given:
[tex]y=f(t)=31(1.02)^t[/tex]
A) We are to write the exponential function in the form
[tex]y=ae^{kt}[/tex]
When t = 1
[tex]f(1)=31(1.02)^1=31(1.02)[/tex]
Also,
[tex]f(1)=ae^{k(1)}=ae^k[/tex]
Equating the two equations and solving for k
[tex]\begin{gathered} 31(1.02)=ae^k \\ \therefore a=31 \\ We\text{ then have,} \\ 1.02=e^k \end{gathered}[/tex]
Apply exponent rules:
[tex]\begin{gathered} k=\ln \mleft(1.02\mright)=0.01980\approx0.0198(4\text{ decimal places)} \\ \therefore k=0.0198 \end{gathered}[/tex]
B) The annual growth rate is,
[tex]\begin{gathered} y=ab^x \\ \text{where,} \\ b=1+r=1.02 \end{gathered}[/tex]
Equating the two expressions together and solving for r
[tex]\begin{gathered} 1+r=1+0.02 \\ r=1+0.02-1=1-1+0.02=0.02=2\% \\ \therefore r=2\% \end{gathered}[/tex]
Hence, the annual growth rate is 2% per year.
C) The continuous growth rate is the constant k which is
[tex]\begin{gathered} k=0.0198=1.98\% \\ \therefore k=1.98\% \end{gathered}[/tex]
Hence, the continuous growth rate is 1.98% per year.
