i am unsure how to solve this ,this is a practice sheet

The model function is
[tex]P(t)=12\times3^{0.498t}[/tex]where t is the time in days and P(t) is the population along the time.
Question a.
Initially, at t=0, the population is
[tex]\begin{gathered} P(0)=12\times3^0 \\ P(0)=12\times1 \\ P(0)=12 \end{gathered}[/tex]that is, there was 12 fruits.
Question b.
After t=6, we have
[tex]\begin{gathered} P(6)=12\times3^{0.498\cdot6} \\ P(6)=12\times3^{2.988} \\ P(6)=12\times26.646 \\ P(6)=319.76 \end{gathered}[/tex]that is, by rounding up, there will be 320 fruits.
Question c.
In this case, we have P(t)=8 000, then we have
[tex]8000=12\times3^{0.498t}[/tex]If we move 12 to the left hand side, we get
[tex]\begin{gathered} \frac{8000}{12}=3^{0.498t} \\ 666.666=3^{0.498t} \end{gathered}[/tex]By applying natural logarithms on both sides, we obtain
[tex]\log 666.666=0.498t\log 3[/tex]then, t is equal to
[tex]t=\frac{\log \text{ 666.666}}{\text{0.498log3}}[/tex]therefore, t is
[tex]\begin{gathered} t=\frac{6.502}{0.498(1.098)} \\ t=\frac{6.502}{0.547} \end{gathered}[/tex]and t is equal to 11.88 days. By rounding up, t=12 days.
Question d.
The number of fruits