Judith puts $5000 into an investment account with interest compounded continuously. which approximate annual rate is needed for the account to grow to $9110 after 30 years?

In the question, continuously should be annually.

Solution:

Applicable formula is;

A = P (1+r) ^n

Where;

A = Total amount after 30 years = $9,110

P = Amount invested = $5,000

r = Annual interest rate in decimals

n = Number of years = 30

Substituting;

9110 = 5000 (1+r) ^30

9110/5000 = (1+r) ^30

1.822 = (1+r) ^30

Taking natural logs on both sides;

ln (1.822) = 30 ln (1+r)

0.5999 = 30 ln (1+r)

0.5999/30 = ln (1+r)

0.019998 = ln (1+r)

Taking exponents on both sides

e^0.019998 = 1+r

1.0202 = 1+r

r = 1.0202 - 1 = 0.0202 = 2.02%

Therefore, annual interest rate should be 2.02%.