compounded interest

**math321** · December 21st 2009, 01:52 PM

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

$A=Pe^{rt}$

**pickslides** · December 21st 2009, 01:58 PM

$20000 = P\times e^{0.073\times 12}$

Find P.

After this find t for $3\times P$

**math321** · December 21st 2009, 03:24 PM

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

**pickslides** · December 21st 2009, 03:31 PM

$20000 = Pe^{0.073\times 12} \implies P = \frac{20000}{e^{0.073\times 12}} \implies P = 8329$

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

Is this what you found?

Spoiler:

**skeeter** · December 21st 2009, 03:33 PM

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

$3P = Pe^{rt}$

divide both sides by P ...

$3 = e^{rt}$

use logs and solve for $t$

**math321** · December 21st 2009, 04:38 PM

this is wat i did
this is how i managed to get a negative number

$20000=8323.93*e^{.073*12}$

**skeeter** · December 21st 2009, 04:52 PM

Originally Posted by math321

this is wat i did
this is how i managed to get a negative number

$20000=8323.93*e^{.073*12}$

first, the original investment is $P = 8328.91$ .

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 ...

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

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

**math321** · December 21st 2009, 04:54 PM

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

**math321** · December 21st 2009, 05:20 PM

for my final answer i got 15.05 years

**HallsofIvy** · December 22nd 2009, 04:50 AM

Originally Posted by math321

this is wat i did
this is how i managed to get a negative number

$20000=8323.93*e^{.073*12}$

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?

**skeeter** · December 22nd 2009, 05:17 AM

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

**math321** · December 22nd 2009, 07:56 PM

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

**pickslides** · December 22nd 2009, 08:02 PM

This should be posted in a separate thread

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

When adding logs you can mulitply

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

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

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

expand and solve...

**Jameson** · December 23rd 2009, 07:30 AM

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.

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