Question

1.

You want to make an investment in a continuously compounding account over a period of 20 years. What interest rate is required for your investment to double in that time period? Round the logarithm value to the nearest hundredth and the answer to the nearest tenth.You want to make an investment in a continuously compounding account over a period of 20 years. What interest rate is required for your investment to double in that time period? Round the logarithm value to the nearest hundredth and the answer to the nearest tenth.


3.5%

28.9%

2.89%

35%



2.

You want to make an investment in a continuously compounding account earning 12.6% interest. How many years will it take for your investment to double in value? Round the natural log value to the nearest thousandth. Round the answer to the nearest year.


5 years

1 years

18 years

6 years


3.

The long jump record in feet at a particular school can be modeled by f(x)=18.4+2.5In(X+1) , where x is the number of years since the school was opened. What is the record for the long jump three years after the school opened? Round the logarithm value to the nearest thousandth. Round the answer to the nearest tenth.


20.1 ft.

21.1 ft.

20.9 ft.

21.9 ft.

Answer 1

Part 1)
`r=3.5\%`

Part 2)
`t=6\ years`

Part 3)
`21.9\ ft`

Explanation:

Part 1) we know that

The formula to calculate continuously compounded interest is equal to

`A=P(e)^(rt)`

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

Let

x -----> amount of money to be invested

2x ---> Final Investment Value

we have

`t=20\ years\\ P=\$x\\A=\$2x\\r=?`

substitute in the formula above

`2x=x(e)^(20r)`

`2=(e)^(20r)`

Apply ln both sides

`ln(2)=ln[(e)^(20r)]`

`ln(2)=(20r)ln(e)`

`ln(2)=(20r)`

`0.693=(20r)`

`r=0.693/20`

`r=0.035`

`r=3.5\%`

Part 2) we know that

The formula to calculate continuously compounded interest is equal to

`A=P(e)^(rt)`

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

Let

x -----> amount of money to be invested

2x ---> Final Investment Value

we have

`t=?\ years\\ P=\$x\\A=\$2x\\r=0.126`

substitute in the formula above

`2x=x(e)^(0.126t)`

`2=(e)^(0.126t)`

Apply ln both sides

`ln(2)=ln[(e)^(0.126t)]`

`ln(2)=(0.126t)ln(e)`

`ln(2)=(0.126t)`

`0.693=(0.126t)`

`t=0.693/0.126`

`t=5.5=6\ years`

Part 3) we have

`f(x)=18.4+2.5ln(x+1)`

where

f(x) -----> is the long jump record in feet

x -----> the number of years since the school was opened

so

For x=3 years

substitute

`f(3)=18.4+2.5ln(3+1)`

`f(3)=18.4+2.5ln(4)`

`f(3)=18.4+2.5(1.386)`

`f(3)=21.9\ ft`