Determine when, to the nearest year, $4,000 invested at 5% per year, compounded daily, will be worth $10,000. yr

18 years

Explanation:

The computation of the time period is shown below:

Given that

Present value = $4,000

Rate of interest = 5 years

N = 365 days

Future value = $10,000

Now as we know that

Future value = Present value × (1 + rate of interest)^number of years

Let us assume the number of years be t

$10,000 = $4,000 × (1 + 0.05 ÷ 365)^365 × t

2.5  = (1 + 0.05 ÷ 365)^365 × t

2.5 = 1.000136986^365 × t

Now take log on both the sides

log 2.5 = 365t  × log 1.000136986

t = 1 ÷ 365[log 2.5 ÷ log 1.000136986]

= 18 years

i guess you could put raise the economic balance

answer makes less frequent trips to the bank and firms make less frequent price changes;