1. Interest Compounded Continuously

Currently has $4679, but know he can get a loan at a lower interest rate if he can put down$5489. If he invests the $4679 in an account that earns 4.1% annually, compounded continuously, how long will it take him to accumulate the$5489?

Can anybody help me with working this problem.

2. Once you've seen it, you will know.

$
4679 \cdot e^{t*0.041}\;=\;5489
$

Solve for 't'.

3. ok thank you for your help...ok still need help on solving it...mine aint working out right?!

4. $4679 \cdot e^{t \times 0.041}\;=\;5489$

$e^{t \times 0.041}\;=\; \frac{ 5489}{4679}$

$ln(e^{t \times 0.041})\;=\; ln(\frac{ 5489}{4679})$

$t \times 0.041\;=\; ln(\frac{ 5489}{4679})$

$t \;=\; \frac {ln(\frac{ 5489}{4679})}{0.041}$

$t \; \approx \; 3,89$ (2 d.p.)

5. Originally Posted by fscc
ok thank you for your help...ok still need help on solving it...mine aint working out right?!
Not good. You should have prerequisites that include simple exponentials and logarithms before getting to this class. Have you been skipping around the curriculum?