# Thread: Natural Log and interest

1. ## Natural Log and interest

Right, I am getting terribly inaccurate answers so I need help:
The question is, "With an investment of $1,500 in an account, your goal is to get a$2,500 return. Interest is compound continuously."
Part 1: How long does it take to reach your goal if the rate is 7.65%?
Part 2: What Interest rate would you need to reach your goal in 6 years?

For part 1 I got 6.68 years, which I'm pretty sure is wrong, and if it isn't then I donno how I got it. I'm totally lost on part 2, I know I'm supposed to use P=ert for both, right? Could anyone walk me through this?

2. Originally Posted by jcthecarpenter
Right, I am getting terribly inaccurate answers so I need help:
The question is, "With an investment of $1,500 in an account, your goal is to get a$2,500 return. Interest is compound continuously."
Part 1: How long does it take to reach your goal if the rate is 7.65%?
Part 2: What Interest rate would you need to reach your goal in 6 years?

For part 1 I got 6.68 years, which I'm pretty sure is wrong, and if it isn't then I donno how I got it. I'm totally lost on part 2, I know I'm supposed to use P=ert for both, right? Could anyone walk me through this?
$A=Pe^{rt}$

Where A = total, P = initial investment, r = the percentage rate in decimal form, and t = time in account.

3. Okay I got that, but for R do I plug in .0765 or 1.0765 or what?
Right:
I would use- 1,500=2,500e^(1.0765*t). This would become 1,500/2,500=e^(1.0765*t). This then becomes 1.6repeating=e^(1.0765*t)
This in turn becomes Ln(1.6repeating)=1.0765*t, correct? If so, would I then solve for Ln1.6 (0.51) and divide that by 1.0765? or instead would I do the inverse root 1.0765 of .51? Either way I get crazy decimals, I am confused.

4. It's 10:30p.m. and my test is tomorrow 2nd period, any quick help? =(

5. Originally Posted by elizsimca
$A=Pe^{rt}$

Where A = total, P = initial investment, r = the percentage rate in decimal form, and t = time in account.
For r, you'd put in 0.0765, So:

$2500 = 1500e^{0.0765t}$

$\frac{5}{3} = e^{0.0765t}$

$ln \frac{5}{3} = 0.0765t$

$t = \frac{ln\frac{5}{3}}{0.0765}$

t = 6.677 years

Your problem was you used 1.0765. A similar method, solving for r rather than t, can be used in part 2.