Question

Sung Lee invests $7,000 at age 18. He hopes the investment will be worth $14,000 when he turns 30. If the interest compounds continuously, approximately what rate of growth will he need to achieve his goal? Round to the nearest tenth of a percent.

Answer 1

Answer:

2.3%

Step-by-step explanation:

For continous compound interest

$A=Pe^{it}$

Where A is final value, P is initial investment, i is interest rate in decimal and t is period.

The time will be 30-18=12 years

Substituting $14000 for A, $7000 for P and 12 years for t

$14000=7000e^{12i}\\e^{12i}=2$

Introducing natural logarithms on both sides then 12i=ln 2

Making i the subject of formula then

i=ln 2 ÷12=0.05776226504666

The percentage rounded off will be 5.7% p.a

Answer 2

The rounded off percentage will be "5.7%".

According to the question,

Investment at the age of 30,

• A = $14,000

Investment at the age of 15,

• P = $7,000

Time,

• t = 30-18

          = 12 years

As we know,

For continuous compound interest:

→      $A = Pe^{it}$

By substituting the values, we get

  $14000=7000e^{12i}$    

     $e^{12i}=2$

By taking "ln", we get

      $12i = ln \ 2$

         $i = \frac{ln \ 2}{12}$

           $=0.05776$

or,

           $= 5.7$ (%) p.a

Thus the above approach is right.

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