+0

# culture starts with 10000 bacteria, and the number doubles every 40 minutes.

0
250
2

culture starts with 10000 bacteria, and the number doubles every 40 minutes.

a. Find a function that models the number of bacteria at time t.

b. Find the number of bacterial after one hour.

c. After how many minutes will there be (50000 +m) bacteria? m=8

d. Sketch a graph of the number of bacterial at time t.

Jun 24, 2021

#1
+1

a) let t be the number of bacteria as its instructing you, A be the starting point, r be the rate of increment and T be time.

lets use the formula t=A\times r^T

T itself can be broken down into 2 components -- the 40 minutes were stated, and another unknown value which would be the time you want it to be; so: $t=\frac{T_m}{40}$

respectively, substitute the values in:

$t=1000\times 2^{\frac{T_m}{40}}$

b) we are working in minutes so one hour is equal to 60 minutes ; just plug that as T_m

$t=1000\times 2^{\frac{60}{40}}$

$t\approx 1000\times 2.828$

$t\approx 2828$

c)  its stated that $t=(50000 +m)$ where $m=8$

$(50000 +8)=1000\times 2^{\frac{T_m}{40}}$ thus,

$50008=1000\times 2^{\frac{T_m}{40}}$

get rid of the base

$2^{\frac{T_m}{40}}=\frac{6251}{125}$

$\frac{T_m}{40}\ln \left(2\right)=\ln \left(\frac{6251}{125}\right)$

$T_m=225.7$

d) use desmos, or try manually graphing it.

Jun 24, 2021
#2
+2

a.      N   = 10000 ( 2)^(t/40)        where N is the number of  bacteria after t  minutes

b.  N  =  10000 (2)^(60/40)    ≈  28284

c.    5000 +  8 =  10000 (2)^(t/40)

50008  =  10000  (2) ^(t/40)

50008  /10000 =  2^(t/40)          take the log of each  side

log   (50008 / 10000)  =  log 2^(t/40)       and we  can write

log (50008 / 10000)   =  (t/40)  log (2)

t =    40 log ( 50008  / 10000) / log (2)  ≈    92.9  minutes

d.  Here's  the graph   ( subbing y for N  and x  for t)

https://www.desmos.com/calculator/t5egnmgcks   Jun 24, 2021
edited by CPhill  Jun 24, 2021