please post the original problem as stated for you to solve ... what you have posted is complete jibberish.
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.
- 100+100X=113.8629436=Ans
- Ans+AnsX=129.6476993=new Ans
- Ans+AnsX=147.6206867=new Ans
- Ans+AnsX=168.0852593=new Ans
- Ans+AnsX=191.386824
Shouldn't line 5 equal 200?
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.
let = initial investmentThe original problem is What constantly compounded interest rate is needed to double your money in two years?
substitute 2 years for and find the required interest rate,
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 .