Question

how long will it take for an investment of $1,000 to double in value if the interest rate is 8.5% per year compounded continuously?

Answer 1

2000 = 1000*e^(.085*t)

can you solve for t by taking log both sides

Answer 2

yes

Answer 3

I got a different answer so I messed up somewhere...

Answer 4

2 = e^(.085t)
ln2 = .085t
t = ln2/.085 = ~ 8.15 years

Answer 5

hmm.. hw much u got

Answer 6

How is the log of 2000 2?

Answer 7

2000 = 1000*e^(.085*t)

dividing both sides by 1000

2 = e^(.085t)

Answer 8

k i got it thanks!

Answer 9

brilliant ! yw ^_^