can you solve this?

Question

As a town gets smaller, the populaion of its high school decreases by 4% each year. The senior class has 160 students now. In how many years will it have 100 students. Write an equation then solve the question

Answer 1

I like to start these problems off by making a recursive relation. That is to say a relation that depends on information that came before.

 

So if we let X_n denote the number of seniors in the high school after n years, then we can write X_n = 0.96*X_n-1

That is to say the current year is 4% less then the previous year.

 

We can then find a closed form solution:

X_n = 0.96*X_n-1 = 0.96*(0.96*X_n-2) = 0.96²*(0.96*X_n-3)= ... = 0.96ⁿ*X₀

Where X₀ is the initial class size 160.

X(n) = 0.96ⁿ*160, now this is not a recursive relation but a function of n.

 

Now lets find out for what n, X(n) = 100

100 = 0.96ⁿ*160

Divide by 160:                    100/160 = 0.96ⁿ

Simplify fraction:                 0.625 = 0.96ⁿ

Take Logarithm:                  log_0.96(.625) = n*log_0.960.96

Logarithm change of base:   log₁₀(.625)/log₁₀(0.96) = n

Plug into a calculator:          n ≈ 11.51349

 

Where n is the number of years. So it will take about 11 and a half years for the class size to be 100.

 

Answer 2

The population depends on the time.  This is because as time passes, population changes.  x is independent variable.  y is dependent variable.

 

Let x = time (in years)

Let y = population

 

The initial population is 160 students.  A year has not passed yet.  Therefore,

 

x=0         y=160

 

After 1 year has passed, the population decreases 4%.  4% is the same as 0.04

 

x=1         y=160 - 160(0.04)=153.6

 

 

So here are your two points:

 

(0, 160) and (1, 153.60)

 

 

Using these two points, create an formula.  Your formula should be an exponential function.

 

y = 160e^kx

 

 

where k is the value we want to solve for.

 

To solve for k, we will use the point (1, 153.60).  Plug in the values of the coordinates into the formula I just created for you.

 

153.60 = 160e^k

 

0.96 = e^k

 

 

Using logarithms to continue to solve for an exponential value,

 

ln(0.96) = ln(e^k)

 

ln(0.96) = k

 

-0.04 = k

 

 

So your function is going to be

 

y = 160e^-0.04x

 

 

Now just plug in y=100 to solve for x from this formula.

Answer 3

This is a question having to do with exponential decay

since the population is decreasing.

 

P(t) = P₀(1-r)^t

P(t) = future or final population amount = 100

P₀ = beginning population = 160

r = rate of decrease = 4% = .04

t = time in years

 

100 = 160(1-.04)^t

Solve for t

 

100 = 160(.96)^t

divide both sides by 160

100/160 = .96^t

5/8 = .96^t

Take log of both sides

log(5/8) = log(.96^t)

log(5/8) = t log .96

divide both sides by log .96

 

[log(5/8)]/log.96 = t

11.5 = t

 

There will be 100 students in 11.5 years