# Thread: compounded interest

1. ## compounded 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}$2.$\displaystyle 20000 = P\times e^{0.073\times 12}$Find P. After this find t for$\displaystyle 3\times P$3. i found P and after that i multiplied it by 3 and solved for t and ended up with a negative number am i suppose to get a negative number because t is years 4.$\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}$Is this what you found? Spoiler:$\displaystyle t\approx 15$5. Originally Posted by math321 i found P and after that i multiplied it by 3 and solved for t and ended up with a negative number am i suppose to get a negative number because t is years would a negative value for t make sense? note you don't need to know the value of P to find how long it takes for the investment, P , to triple$\displaystyle 3P = Pe^{rt}$divide both sides by P ...$\displaystyle 3 = e^{rt}$use logs and solve for$\displaystyle t$6. this is wat i did this is how i managed to get a negative number$\displaystyle 20000=8323.93*e^{.073*12}$7. Originally Posted by math321 this is wat i did this is how i managed to get a negative number$\displaystyle 20000=8323.93*e^{.073*12}$first, the original investment is$\displaystyle P = 8328.91$. now, read the original question again ... 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
the equation should be ...

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

once again, note that the value of P will cancel, leaving you with the equation $\displaystyle 3 = e^{0.073t}$

8. or i now see
wat i did wrong
when im finish ill get back to you to see if it is correct

9. for my final answer i got 15.05 years

10. Originally Posted by math321
this is wat i did
this is how i managed to get a negative number

$\displaystyle 20000=8323.93*e^{.073*12}$
I have no idea what you mean by "this is how i managed to get a negative number". What you show is an equation (which is only very approximately true by the way- the right hand side is about 19988). What did you do to get a negative number?

11. Originally Posted by HallsofIvy
I have no idea what you mean by "this is how i managed to get a negative number". What you show is an equation (which is only very approximately true by the way- the right hand side is about 19988). What did you do to get a negative number?
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

12. ## logs problem

log2 (x+1) + log2 (x-1) = 3 ..... log to base 2

13. 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...

14. Originally Posted by skeeter
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
Way to be on the lookout skeeter. This sort of thing is common I bet but in cases like these can be very aggravating as the OP has no intention of showing any effort.