# Thread: Quick Algebra question (logs and stuff)

1. ## 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.

2. You used 4.5 instead of .045

3. 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.
$4.5\text{ }\text{percent}=\frac{4.5}{100}=.045$

4. aaaaaah!

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

6. 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!

7. EDIT see next post

8. 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?

9. 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:

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

10. eh im confused

11. 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

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

12. 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

$\frac{\ln(\frac{3}{5})}{5}=\frac{\ln\bigg(\bigg(\f rac{5}{3}\bigg)^{-1}\bigg)}{5}=\frac{-1\ln(\frac{5}{3})}{5}$

13. oh, so both come out to the same thing?

14. gah, back to the question: