Thread: What interest rate needed to double money in two years?

e^{5r}=2
5r ln e=ln(2)
r=ln(2)/5=.1386294361 (store in variable X)

Using a calculator where Ans refers to the previous answer.

1. 100+100X=113.8629436=Ans
2. Ans+AnsX=129.6476993=new Ans
3. Ans+AnsX=147.6206867=new Ans
4. Ans+AnsX=168.0852593=new Ans
5. Ans+AnsX=191.386824

Shouldn't line 5 equal 200?

please post the original problem as stated for you to solve ... what you have posted is complete jibberish.

Sorry.

The original problem is What constantly compounded interest rate is needed to double your money in two years?

I setup the problem:
2=e^{5r}

I solved the problem:
5r ln e=ln(2)
r=ln(2)/5=.1386294361

And got an interest rate of approximately 13.86%.

Then I did a check, by starting with $100 and adding 13.86% times 100 to it.
100+100(.1386294361)=113.8629436

I took that answer and added 13.86% to it.
(Ans)+(Ans)(.1386294361)=129.6476993

Then I did that three more times.
(Ans)+(Ans)(.1386294361)=147.6206867
(Ans)+(Ans)(.1386294361)=168.0852593
(Ans)+(Ans)(.1386294361)=191.386824

Shouldn't I have $200 by the fifth time? Or I did the first part of the problem wrong.

Thanks.

Why use 5 r? What is the purpose of the 5?

The original problem is What constantly compounded interest rate is needed to double your money in two years?

let = initial investment









substitute 2 years for and find the required interest rate,

Originally Posted by SammyS

Why use 5 r? What is the purpose of the 5?

Sloppy title. Five years.

Originally Posted by skeeter

let = initial investment









substitute 2 years for and find the required interest rate,

That's what I did. But I wrote the title wrong. The correct time frame is 5 years.

But then, when I apply that for five years, I get $191.39. Shouldn't I get $200?

Originally Posted by mathDad

Sloppy title. Five years.


That's what I did. But I wrote the title wrong. The correct time frame is 5 years.

But then, when I apply that for five years, I get $191.39. Shouldn't I get $200?

I'm guessing that you are using $100 as the initial investment?



using this interest rate (and not rounding it)



if you do round to 13.86% , you'll still get very close (199.97)

Originally Posted by mathDad

e^{5r}=2
5r ln e=ln(2)
r=ln(2)/5=.1386294361 (store in variable X)

Using a calculator where Ans refers to the previous answer.

1. 100+100X=113.8629436=Ans
2. Ans+AnsX=129.6476993=new Ans
3. Ans+AnsX=147.6206867=new Ans
4. Ans+AnsX=168.0852593=new Ans
5. Ans+AnsX=191.386824

Shouldn't line 5 equal 200?

Why should it? (and why are you using "5" for t when the problem asks about 2 years) refers to "continuous compounding". In the latter part of your conputation you are compounding annually (for 5 years, not 2).

Compounding annually, with interest rate r, will give, in n years, . The interest rate necessary to double the amount in 5 years (I assume that is what you meant), compounding annually, is given by solving . Taking the logarithm of both sides, so that . Then [tex]1+ r= e^{0.1386}= 1.1486983550[tex] and so .