How many years will it take for the account to reach $18,600? Round your answer to the nearest hundredth

it depends how the interest is calculated, but there's not much of a difference

assuming its continuously compouned, you use this formula: A (t) = Pe^ (rt), where A is the final amount, P is the initial investment, r is the interest, and t is the time in years

you want to find t such that A (t) = 18,600 so 18,600=1000e^ (.0675t)

you need to use logarithm to figure it out, take the natural log of both sides

the following properties will come into use:

ln (a*b) = ln (a) + ln (b)

ln (a^b) = bln (a)

ln (e) = 1

taking the natural log

ln (18,600) = ln (1000e^ (.0675t))

ln (18,600) = ln (1000) + ln (e^.0675t)

ln (18600) = ln (1000) +.0675t

now solve for t: t = (ln (18600) - ln (1000)) /.0675

t=43.31