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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.

 

 

 

please help me

 Jun 24, 2021
 #1
avatar+171 
+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
avatar+129845 
+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

 

 

cool cool cool

 Jun 24, 2021
edited by CPhill  Jun 24, 2021

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