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

Where is the diagram?