Achemical 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.)

a)  

b)   grams

c)   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.

Multiply both sides by :

Divided both sides by :

Integrate both sides:

The equivalent exponential form is:

Using law of exponents you can write this as:

is just a positive constant that I'm going to replace with K:

Applying the symmetric property of equality:

Applying the commutative property of multiplication:

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

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

b)

We are given at .

We are asked to find how much of the chemical substance, N, remains after 2 days.  So we replace t with 2 in :

Put into calculator:

(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:

Divide both sides by 800:

Reduce the fraction:

Convert to logarithmic form:

Divide both sides by -0.069:

Input into calculator:

answer: all the values are within because it has a pattern that repeats from -6 to 4.

good luck!