Exponential and logarithmic functions help

Question

The population of New Bedford is declining. After n years the population can be predicted using the function P(n)=13,700e^-0.013n what will the population be in 10 years? If it continues to decline at this rate how long will it take for the population of New Bedford to drop to 10,000? What interest rate is needed for an and investment of $100,000 to double in 30 years? assume continues compounding

Answer 1

Hi Alex,

 

 The population of New Bedford is declining. After n years the population can be predicted using the function P(n)=13,700e^-0.013n what will the population be in 10 years? If it continues to decline at this rate how long will it take for the population of New Bedford to drop to 10,000? What interest rate is needed for an and investment of $100,000 to double in 30 years? assume continues compounding

 

 

1.  After n years the population can be predicted using the function P(n)=13,700e^-0.013n what will the population be in 10 years?

 

For this part, substitute in 10 for n.

 

P(n)=13,700e^-0.013n

P(10)=13,700e^-0.013(10)    Use calculator

P(10)=12029.9074    Since living populations can't have a part of a body, the answers are usually rounded down to the nearest whole number.  Therefore, In 10 years the population will be 12029.

 

2.  If it continues to decline at this rate how long will it take for the population of New Bedford to drop to 10,000?

 

 

P(n)=13,700e^-0.013n

10,000=13,700e^-0.013n     divide both sides by 13700

100/137=e^-0.013n             Since this is base e, take the natural logarithm of both sides

ln(100/137)=ln e^-0.013n      Since ln e^x=x, the ln e^-0.013n=-0.013n

ln(100/137)=-0.013n          Divide by -0.013

[ln(100/137)]/-0.013=n       Use calcutator

n=24.2162                        Rounded to 4 decimal places

 

3.  What interest rate is needed for an and investment of $100,000 to double in 30 years? assume continues compounding

 

P(n)=13,700e^-0.013n    The interest rate in this equation is the amount is decreasing by 1.3%

                                  and the initial deposit is $13,700

 

Now, you are give the time (30 years) and the initial amount of investment ($100,000).  In this case, the rate will be positive.

 

If an investment of $100,000 doubles it will be $200,000.

Let r=interest rate

 

      200,000=100,000e^30r                Divide both sides by 100,000

                2=e^30r                            Since the base is e, take the natural logarithm of both sides

            ln 2 = ln e^30r

           ln 2  = 30r                            Divide both sides by 30

           (ln 2)/30=r                            Use calculator

          0.0231=r

 

ANSWER:  r=0.0231; The rate needs to be 2.31% for an initial deposit of $100,000 to double in 30 years.

 

Hope this helps you to understand the process.