Continuous Growth

[paperangel]

PROBLEM:
An investment of $5,000.00 averages an interest rate of 8.5% when invested in a stock based mutual fund. Determine the doubling time for this investment if the interest is compounded continuously.

THE ANSWER
8.15 years

MY WORK:
It's continuous growth so it uses A=Pe^(rt)
r = .85
P = 5,000
T = what we're looking for.
A = 10,000

10,000 = 5000e^(.85t)

Then I put this into my calculator(the x is the variable, not an alphalock): 5000e^(.85x)
And I am getting crazy answers. What am I doing wrong?

 

[stapel]

10,000 = 5000e^(.85t)
Then I put this into my calculator...And I am getting crazy answers. What am I doing wrong?

Without knowing the specific keystrokes being used (on what model of calculator), what you are trying to accomplish, or what you mean by "crazy answers", it is difficult to analyze whatever you might be doing or getting. Sorry.

Eliz.

 

[Unco]

r = 0.085

 

[paperangel]

Without knowing the specific keystrokes being used (on what model of calculator), what you are trying to accomplish, or what you mean by "crazy answers", it is difficult to analyze whatever you might be doing or getting. Sorry.

No. It is my fault. I didn't mean to be so vague. I forget that people can't read my mind, sometimes. :roll:

I'm using a Texas Instrument graphing calculator model: TI-83.

The exact keystrokes I am inserting are:

5000e^(.085X)

The X isn't from the alphabet lock, it is a variable.

I am getting an overflow error from my calculator.

Obviously I am doing something wrong - do I have to put in something different because I'm looking from how long it will take for it to double? Should I put in 10000 and work backwards (I'm not sure how I would do that)?

 

[paperangel]

r = 0.085

Thanks, I noticed that as I read my own post. Silly mistakes - that is what a lot of this is. :roll: You're very kind.

 

[stapel]

The exact keystrokes I am inserting are:
5000e^(.085X)
The X isn't from the alphabet lock, it is a variable.

The calculator is evaluating the expression at whatever value was last used (and thus stored) in the system variable "X". What value have you stored in this system variable?

And why are you evaluating the expression for some predetermined value of the calculator variable X, instead of solving the posted equation to find the value of equation variable x? Am I misunderstanding the exercise...?

Eliz.

 

[paperangel]

The exact keystrokes I am inserting are:
5000e^(.085X)
The X isn't from the alphabet lock, it is a variable.

The calculator is evaluating the expression at whatever value was last used (and thus stored) in the system variable "X". What value have you stored in this system variable?

And why are you evaluating the expression for some predetermined value of the calculator variable X, instead of solving the posted equation to find the value of equation variable x? Am I misunderstanding the exercise...?

Eliz.

No, I'm just doing a terrible job at explaining.

The "X" that I am using is not the one where you can store a desired value into it.

I am trying to figure out how long it will take me to get from 5000 dollars to 10000 dollars with a continously compounded interest rate of 8.5%.

I loaned out my notes to a classmate from this section and they dropped before they gave them back. -_-

 

[Gene]

You ARE using the x that was last set in your calculations. There is only one x. You can get it by ([ ] is a calc key)
[ALPHA] [X] or
[X,T,theta,n]
It's the same one. You can
8.2[STO->]X
then put in your equation
to check whether the answer is 8.2
The correct way is
10000=5000e^(.085X)
2=e^(.085X)
ln(2)=.085X
X=ln(2)/.085
Put the ln(2)/.085 into your calculator and hit enter.
OR
[Y=]
Enter your equation in Y₁
[WINDOW]
Xmin=0
Xmax=10
Xscl=1
Ymin=0
Ymax=12000
Yscl=1000
[GRAPH]
[TRACE]
[->] till y=10000
read X=?

BTW you can check a doubling time by
the rule of 70
70/I% = doubling time (about)
70/8.5 = 8.235 yrs

 

[paperangel]

Wow. I never remember the most obvious things. You are my hero. Thankyouthankyouthankyouthankyouthankyou!