MATH SOLVE

4 months ago

Q:
# 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.β

Accepted Solution

A:

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 tStep 1: find the rate Β [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]