# Quick Algebra question (logs and stuff)

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• May 17th 2008, 11:44 AM
reynardin
Quick Algebra question (logs and stuff)
I'm having trouble with this word problem:

Assume $150 is deposited in a savings account. If the interest rate is 4.5% compounded continuously, after how many years will the investment be worth$450? (Round your answer to the nearest month.)

I start off with
$y=ae^{rt}$
$450=150e^{4.5t}$
$3=e^{4.5t}$
$\ln(3)=4.5t\ln(e)$

after this, my teacher says i did something wrong, but i can't figure out where i did it wrong.
• May 17th 2008, 11:46 AM
galactus
You used 4.5 instead of .045
• May 17th 2008, 11:48 AM
Mathstud28
Quote:

Originally Posted by reynardin
I'm having trouble with this word problem:

Assume $150 is deposited in a savings account. If the interest rate is 4.5% compounded continuously, after how many years will the investment be worth$450? (Round your answer to the nearest month.)

I start off with
$y=ae^{rt}$
$450=150e^{4.5t}$
$3=e^{4.5t}$
$\ln(3)=4.5t\ln(e)$

after this, my teacher says i did something wrong, but i can't figure out where i did it wrong.

$4.5\text{ }\text{percent}=\frac{4.5}{100}=.045$
• May 17th 2008, 12:01 PM
reynardin
• May 17th 2008, 12:03 PM
reynardin
oh yay! mathstud28 again! =D
I still got 2 more problems from yesterday that I don't get lol.
• May 17th 2008, 12:11 PM
Mathstud28
Quote:

Originally Posted by reynardin
oh yay! mathstud28 again! =D
I still got 2 more problems from yesterday that I don't get lol.

Ok go! (Rofl)
• May 17th 2008, 12:37 PM
reynardin
EDIT see next post
• May 17th 2008, 12:39 PM
reynardin
ok;
In 5 years, radioactivity reduces the msas of a 100-gram sample of an element to 60 grams. for this element, find the number of years it would take for the original sample to be reduces to half its original mass.

im using the equation $y=ae^{-kt}$ to start.
$60=100e^{-.5t}$
$.6 = e^{-.5t}$

good so far? or not?
• May 17th 2008, 12:48 PM
galactus
Using the initial condition.

y(0)=100, y(5)=60

$60=100e^{5k}$

Solving for $k=\frac{-ln(\frac{5}{3})}{5}\approx{-.102165124753}$

So, now solve the following for t:

$50=100e^{\frac{-ln(5/3)}{5}t}$
• May 17th 2008, 12:49 PM
reynardin
eh im confused
• May 17th 2008, 12:53 PM
Mathstud28
Quote:

Originally Posted by galactus
Using the initial condition.

y(0)=100, y(5)=60

$60=100e^{5k}$

Solving for $k=\frac{-ln(\frac{5}{3})}{5}\approx{-.102165124753}$

So, now solve the following for t:

$50=100e^{\frac{-ln(5/3)}{5}t}$

We know that when t=0 that y=100

setting this up we get

$100=Ce^{k\cdot{0}}=Ce^0=C$

So C=100

Next we have that t=5 y=60

Setting this up we get

$60=100e^{k\cdot{50}}\Rightarrow{\frac{3}{5}=e^{5k} }\Rightarrow{\ln\bigg(\frac{3}{5}\bigg)=5k}\Righta rrow{k=\frac{\ln\bigg(\frac{3}{5}\bigg)}{5}}$
• May 17th 2008, 12:55 PM
Mathstud28
Quote:

Originally Posted by Mathstud28
We know that when t=0 that y=100

setting this up we get

$100=Ce^{k\cdot{0}}=Ce^0=C$

So C=100

Next we have that t=5 y=60

Setting this up we get

$60=100e^{k\cdot{50}}\Rightarrow{\frac{3}{5}=e^{5k} }\Rightarrow{\ln\bigg(\frac{3}{5}\bigg)=5k}\Righta rrow{k=\frac{\ln\bigg(\frac{3}{5}\bigg)}{5}}$

O

Just so our conflicting answers dont confuse you

$\frac{\ln(\frac{3}{5})}{5}=\frac{\ln\bigg(\bigg(\f rac{5}{3}\bigg)^{-1}\bigg)}{5}=\frac{-1\ln(\frac{5}{3})}{5}$
• May 17th 2008, 01:19 PM
reynardin
oh, so both come out to the same thing?
• May 17th 2008, 01:22 PM
reynardin
gah, back to the question: