The populations both colony have the same size after 2.156 hours.
Given that,
The size P of the colony, measured in grams, to decrease,
According to the exponential decay model dP/dt=−0.4P,
Where the time t is measured in hours.
The size Q of a second colony of microorganisms, also measured in grams, decreases at the constant rate of 1 gram per hour
According to the linear model dQ/dt=−1.
If at time t=0 the first colony has size P(0)=2 and the second colony has size Q(0)=3,
We have to determine,
What time will both colonies have the same size.
According to the question,
[tex]\dfrac{dp}{dt} = 0.4p[/tex]
Separate the variables and integrate,
[tex]\dfrac{dp}{p} = 0.4 dt\\\\integrate\ both \ the \ sides ,\\\\\int\dfrac{dp}{p} = \int0.4dt\\\\ln|p| = 0.4t+c\\\\Taking \ exponential \ both \ sides\\\\P = c.e^{(0.4t)}[/tex]
Using initial condition to find C.
[tex]2 = c^{0} \\\\c = 2[/tex]
Substitute c = 2 in the equation,
[tex]P = 2.e^{(0.4t)}[/tex]
Again, integrate the second function,
[tex]\dfrac{dQ}{dt} = -1\\\\On \ integrating \ both \ the \ sides\\\\\int dQ = \int -1.dt\\\\Q = -t +c \\[/tex]
Use initial condition to find C.
[tex]3 = 0+c[/tex]
Then ,
[tex]Q = -t +3[/tex]
On comparing both the equation to get the value of t,
[tex]2. e^{(0.4t)} = -t+3\\\\t = 2.156[/tex]
Hence, The populations both colony have the same size after 2.156 hours.
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