The solution is Option A.
The exponential growth model for the number of bacteria in the sample after x hours is [tex]P ( x ) = 200 e ^{(\frac{ln 4}{8})x }[/tex]
What is exponential growth factor?
The exponential growth or decay formula is given by
x ( t ) = x₀ × ( 1 + r )ⁿ
x ( t ) is the value at time t
x₀ is the initial value at time t = 0.
r is the growth rate when r>0 or decay rate when r<0, in percent
t is the time in discrete intervals and selected time units
Given data ,
Let the equation for the number of bacteria in the sample after x hours = P
The value of the equation P ( x ) is given by
After 4 hours , the sample contained 400 bacteria
So ,
when x = 4
[tex]P ( 4 ) = ae ^{4b }=400[/tex] be equation (1)
After 12 hours , the sample contained 1600 bacteria
And , when x = 12
[tex]P ( 12 ) = ae ^{12b }=1600[/tex] be equation (2)
Divide equation (2) by equation (1) , we get
[tex]\frac{P ( 12 )}{P ( 4 )} = \frac{ae ^{12b }}{ae ^{4b }} = \frac{1600}{400}[/tex]
On simplifying the equation , we get
e¹²ᵇ⁻⁴ᵇ = 4
e⁸ᵇ = 4
Taking logarithm on both sides of the equation , we get
8b = ln (4)
Divide by 8 on both sides , we get
b = ln (4) / 8
Substituting the value for b in equation (1) , we get
[tex]ae ^{4*ln (4) / 8 }=400[/tex]
[tex]ae ^{ln (4) / 2}=400[/tex]
On simplifying the equation , we get
a x 2 = 400
Divide by 2 on both sides of the equation , we get
a = 200
Therefore , the exponential growth equation is given by
Substitute the values of a and b in the equation , we get
[tex]P ( x ) = 200 e ^{(\frac{ln 4}{8})x }[/tex]
Hence , The exponential growth model for the number of bacteria in the sample after x hours is [tex]P ( x ) = 200 e ^{(\frac{ln 4}{8})x }[/tex]
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