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A particular bacteria population on an athlete's foot doubles every 3 days. Determine an expression for the number of bacteria N after T days, given the initial amount is 40 bacteria.​

User Dimona
by
4.7k points

2 Answers

5 votes

Answer:


\boxed{N = 40(2)^{(T)/(3)}}

Explanation:

The growth of bacteria is an exponential function. The equation has the general form


f(x) = ab^(x)

Using the variables N and T, we can rewrite the equation as


N = ab^(T)

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.


\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)^{(1)/(3)} &\text{Took the cube root of each side}\\\end{array}\\\\\text{Thus, the explicit equation is } N = 40 \left (2^{(1)/(3)\right )^(T)}} \text{ or}\\\\\boxed{\mathbf{N = 40(2)^{(T)/(3)}}}

User Prasad Pawar
by
5.0k points
3 votes

Answer:
\bold{N=40e^{\bigg((ln2)/(3)\bigg)T}}

Explanation:

The exponential growth formula is:


A=Pe^(rt)\\\bullet A=final\ amount\\\bullet P=initial\ amount\\\bullet r=rate\ of\ growth\\\bullet t=time

NOTE: This problem is asking to use N instead of A and T instead of t

Step 1: find the rate


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{(ln2)/(3)=r}\quad \qquad \leftarrow (divided\ both\ sides\ by\ 3)

Step 2: input the rate to find N


N=Pe^(rT)\\\\\bullet P=40\\\\\bullet r=(ln2)/(3)\\\qquad \implies \qquad \boxed{N=40e^{\bigg((ln2)/(3)\bigg)T}}

User WindsurferOak
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