The given exponential function is
[tex]Q=11e^{-0.13t}[/tex]
The form of the exponential continuous function is
[tex]y=ae^{rt}[/tex]
a is the initial amount (value y at t = 0)
r is the rate of growth/decay in decimal
Compare the given function by the form
[tex]a=11[/tex][tex]\begin{gathered} r=0.13\rightarrow decay \\ r=0.13\times100\text{ \%} \\ r=13\text{ \%} \end{gathered}[/tex]
a)
The value of Q at t = 0 is 11 and the decay rate is 13%
The initial value of Q is 11
The continuous decay rate is 13%
From the graph
To find the value of t when Q = 2
Look at the vertical axis Q and go to the scale of 2
Move horizontally from 2 until you cut the graph
Go down to read the value of t
The value of t is about 13
b)
At
Q = 2
t = 13
c)
Now, we will substitute Q in the function by 2
[tex]2=11e^{-0.13t}[/tex]
Divide both sides by 11
[tex]\frac{2}{11}=e^{-0.13t}[/tex]
Insert ln on both sides
[tex]ln(\frac{2}{11})=lne^{-0.13t}[/tex]
Use the rule
[tex]lne^n=n[/tex][tex]lne^{-0.13t}=-0.13t[/tex]
Substitute it in the equation
[tex]ln(\frac{2}{11})=-0.13t[/tex]
Divide both sides by -0.13
[tex]\begin{gathered} \frac{ln(\frac{2}{11})}{-0.13}=t \\ \\ 13.11344686=t \end{gathered}[/tex]
At
Q = 2
t = 13.11344686
