Answer:
The answer is "26179.4".
Step-by-step explanation:
Assume year 2000 as t, that is t =0.
Formula:
[tex]A= A_0e^{rt}[/tex]
Where,
[tex]A_0 = \ initial \ pop \\\\r= \ rate \ in \ decimal \\\\t= \ time \ in \ year[/tex]
for doubling time,
[tex]r = \frac{log (2)}{t} \\[/tex]
[tex]r = \frac{\log (2)}{ 40} \\\\r= \frac{0.301}{40}\\\\r= 0.007[/tex]
Given value:
[tex]A = A_0e^{rt} \\\\[/tex]
[tex]A_0 = 13000[/tex]
[tex]t= 40 \ years[/tex]
when year is 2000, t=0 so, year is 2100 year as t = 100.
[tex]A = 13000 \times e^{et}\\\\A = 13000 \times e^{e \times t}\\\\A = 13000 \times e^{0.007 \times 100}\\\\A = 13000 \times e^{0.7}\\\\A= 13000\times 2.0138\\\\A = 26179.4[/tex]
