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%.

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Step-by-step explanation: