what sum of money must be invested at 7.3% interest compounded continuously to yield $20,000 in 12 years

how long will it take for the investment to triple in value

$\displaystyle A=Pe^{rt}$

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- Dec 21st 2009, 01:52 PMmath321compounded interest
what sum of money must be invested at 7.3% interest compounded continuously to yield $20,000 in 12 years

how long will it take for the investment to triple in value

$\displaystyle A=Pe^{rt}$ - Dec 21st 2009, 01:58 PMpickslides
$\displaystyle 20000 = P\times e^{0.073\times 12}$

Find P.

After this find t for $\displaystyle 3\times P$ - Dec 21st 2009, 03:24 PMmath321
i found P and after that i multiplied it by 3 and solved for t and ended up with a negative number am i suppose to get a negative number because t is years

- Dec 21st 2009, 03:31 PMpickslides
$\displaystyle 20000 = Pe^{0.073\times 12} \implies P = \frac{20000}{e^{0.073\times 12}} \implies P = 8329$

Now solve for $\displaystyle t $ where $\displaystyle 3\times 8329 = 8329 e^{0.073t}$

Is this what you found?

__Spoiler__: - Dec 21st 2009, 03:33 PMskeeter
would a negative value for t make sense?

note you don't need to know the value of P to find how long it takes for the investment, P , to triple

$\displaystyle 3P = Pe^{rt}$

divide both sides by P ...

$\displaystyle 3 = e^{rt}$

use logs and solve for $\displaystyle t$ - Dec 21st 2009, 04:38 PMmath321
this is wat i did

this is how i managed to get a negative number

$\displaystyle 20000=8323.93*e^{.073*12}$ - Dec 21st 2009, 04:52 PMskeeter
first, the original investment is $\displaystyle P = 8328.91$.

now, read the original question again ...

Quote:

what sum of money must be invested at 7.3% interest compounded continuously to yield $20,000 in 12 years

how long will it take for the**investment to triple**in value

$\displaystyle 3(\textcolor{red}{8328.91})=\textcolor{red}{8328.9 1} \cdot e^{.073t}$

once again, note that the value of P will cancel, leaving you with the equation $\displaystyle 3 = e^{0.073t}$ - Dec 21st 2009, 04:54 PMmath321
or i now see

wat i did wrong

when im finish ill get back to you to see if it is correct - Dec 21st 2009, 05:20 PMmath321
for my final answer i got 15.05 years

- Dec 22nd 2009, 04:50 AMHallsofIvy
- Dec 22nd 2009, 05:17 AMskeeter
who knows?

the OP has been bouncing back and forth between sites to find a solution ...

S.O.S. Mathematics CyberBoard :: View topic - compounded interest - Dec 22nd 2009, 07:56 PMmath321logs problem
log2 (x+1) + log2 (x-1) = 3 ..... log to base 2

- Dec 22nd 2009, 08:02 PMpickslides
This should be posted in a separate thread

$\displaystyle \log_2 (x+1) + \log_2 (x-1) = 3$

When adding logs you can mulitply

$\displaystyle \log_2 ((x+1)(x-1)) = 3$

$\displaystyle (x+1)(x-1) = 2^3$

$\displaystyle (x+1)(x-1) = 8$

expand and solve... - Dec 23rd 2009, 07:30 AMJameson