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A chemical substance has a decay rate of 6.9% per day. The rate of change of an amount N of the chemical is given by the equation dN/dt = - 0.069 N where t is the number of days since decay began. Complete parts a) through c). a) Let N_0 represent the amount of the chemical substance at t = 0. Find the exponential function that models the situation. N(t) = N_0 b) Suppose that 800 g of the chemical substance is present at t = 0. How much will remain after 2 days? After 2 days, g will remain. After 2 days, g will remain. (Round to the nearest whole number as needed.) c) After how many days will half of the 800 g of the chemical substance remain? After days, half of the chemical substance will remain. (Round to one decimal place as needed.)

Respuesta :

Answer:

a)  [tex]N=N_0e^{-0.069t}[/tex]

b)  [tex]N=696.9[/tex] grams

c)  [tex]t=10[/tex] days

Step-by-step explanation:

a)

We are going to use separation of variables to solve.

Get all your t's to one side and your N's to opposing side.

[tex]\frac{dN}{dt}=-0.069N[/tex]

Multiply both sides by [tex]dt[/tex]:

[tex]dN=-0.069N dt[/tex]

Divided both sides by [tex]N[/tex]:

[tex]\frac{dN}{N}=-0.069 dt[/tex]

Integrate both sides:

[tex]\ln|N|=-0.069t+C[/tex]

The equivalent exponential form is:

[tex]e^{-0.069t+C}=N[/tex]

Using law of exponents you can write this as:

[tex]e^{-0.069t}e^C=N[/tex]

[tex]e^C[/tex] is just a positive constant that I'm going to replace with K:

[tex]e^{-0.069t}K=N[/tex]

Applying the symmetric property of equality:

[tex]N=e^{-0.069t}K[/tex]

Applying the commutative property of multiplication:

[tex]N=Ke^{-0.069t}[/tex]

K actually represents the initial amount of chemical substance since when plugging in 0 for t you get K for N, like so:

[tex]N=Ke^{-0.069 \cdot 0}[/tex]

[tex]N=Ke^{0}[/tex]

[tex]N=K(1)[/tex]

[tex]N=K[/tex]

We are given at time 0 the amount of chemical substance,N, is K. They want us to represent this value with [tex]N_0[/tex] instead. So the exponential equation is:

[tex]N=N_0e^{-0.069t}[/tex]

b)

We are given [tex]N_0=800[/tex] at [tex]t=0[/tex].

We are asked to find how much of the chemical substance, N, remains after 2 days.  So we replace t with 2 in [tex]N=800e^{-0.069t}[/tex]:

[tex]N=800e^{-0.069 \cdot 2}[/tex]

Put into calculator:

[tex]N=696.9[/tex] (this was rounded to the nearest tenths)

c)  

The last part is asking for how many days will it take a initial 800 grams to go down to half of 800 grams.

We need to see the following equation:

[tex]\frac{1}{2}(800)=800e^{-0.069t}[/tex]

[tex]400=800e^{-0.069t}[/tex]

Divide both sides by 800:

[tex]\frac{400}{800}=e^{-0.069t}[/tex]

Reduce the fraction:

[tex]\frac{1}{2}=e^{-0.069t}[/tex]

Convert to logarithmic form:

[tex]\ln(\frac{1}{2})=-0.069t[/tex]

Divide both sides by -0.069:

[tex]\frac{\ln(\frac{1}{2})}{-0.069}=t[/tex]

Input into calculator:

[tex]10.0=t[/tex]

[tex]t=10.0[/tex]

[tex]t=10[/tex]

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