Answer: [tex]\bold{N=40e^{\bigg(\dfrac{ln2}{3}\bigg)T}}[/tex]
Step-by-step explanation:
The exponential growth formula is:
[tex]A=Pe^{rt}\\\bullet A=final\ amount\\\bullet P=initial\ amount\\\bullet r=rate\ of\ growth\\\bullet t=time[/tex]
NOTE: This problem is asking to use N instead of A and T instead of t
Step 1: find the rate
[tex]N=Pe^{rT}\\2P=Pe^{r\cdot 3}\quad \leftarrow(Initial\ population\ doubled\ N=2P, T=3\ days)\\2=e^{3r}\quad \qquad \leftarrow (divided\ both\ sides\ by\ P)\\ln\ 2=ln\ e^{3r}\quad \leftarrow(applied\ ln\ to\ both\ sides)\\ln\ 2=3r\quad \qquad \leftarrow (ln\ e\ cancelled\ out)\\\boxed{\dfrac{ln2}{3}=r}\quad \qquad \leftarrow (divided\ both\ sides\ by\ 3)[/tex]
Step 2: input the rate to find N
[tex]N=Pe^{rT}\\\\\bullet P=40\\\\\bullet r=\dfrac{ln2}{3}\\\qquad \implies \qquad \boxed{N=40e^{\bigg(\dfrac{ln2}{3}\bigg)T}}[/tex]
Answer:
[tex]\boxed{N = 40(2)^{\frac{T}{3}}}[/tex]
Step-by-step explanation:
The growth of bacteria is an exponential function. The equation has the general form
[tex]f(x) = ab^{x}[/tex]
Using the variables N and T, we can rewrite the equation as
[tex]N = ab^{T}[/tex]
We have two conditions:
(1) There are 40 bacteria at T = 0
(2) There are 80 bacteria at T = 3.
Insert these values into the equation.
[tex]\begin{array}{rrcll}(1)&40& = & a(b)^{0} & \\(2)&80 & = & a(b)^{3} & \\(3)& a & = & 40 & \text{Simplified (1)}\\ &80 & = & 40(b)^{3} & \text{Substituted (3) into (2)}\\ & b^{3} & = & 2 & \text{Divided each side by 40}\\ & b & = & (2)^{\frac{1}{3}} &\text{Took the cube root of each side}\\\end{array}\\\\\text{Thus, the explicit equation is } N = 40 \left (2^{\frac{1}{3}\right )^{T}}} \text{ or}\\\\\boxed{\mathbf{N = 40(2)^{\frac{T}{3}}}}[/tex]
