# Math Help - What interest rate needed to double money in two years?

1. ## 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?

2. ## Re: What interest rate needed to double money in two years?

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

3. ## Re: What interest rate needed to double money in two years?

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.

4. ## Re: What interest rate needed to double money in two years?

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

5. ## Re: What interest rate needed to double money in two years?

The original problem is What constantly compounded interest rate is needed to double your money in two years?
let $y_0$ = initial investment

$2y_0 = y_0 e^{rt}$

$2 = e^{rt}$

$\ln{2} = rt$

$\frac{\ln{2}}{t} = r$

substitute 2 years for $t$ and find the required interest rate, $r$

6. ## Re: What interest rate needed to double money in two years?

Originally Posted by SammyS
Why use 5 r? What is the purpose of the 5?
Sloppy title. Five years.

Originally Posted by skeeter
let $y_0$ = initial investment

$2y_0 = y_0 e^{rt}$

$2 = e^{rt}$

$\ln{2} = rt$

$\frac{\ln{2}}{t} = r$

substitute 2 years for $t$ and find the required interest rate, $r$
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?

7. ## Re: What interest rate needed to double money in two years?

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?

$r = \frac{\ln{2}}{5} = 0.1386...$

using this interest rate (and not rounding it)

$100e^{r \cdot 5} = 200$

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

8. ## Re: 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?
Why should it? $e^{rt}$ (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, $A= P(1+ r)^n$. The interest rate necessary to double the amount in 5 years (I assume that is what you meant), compounding annually, is given by solving $(1+ r)^5= 2$. Taking the logarithm of both sides, $log(1+r)^5= 5log(1+ r)= log(2)$ so that $log(1+r)= log(2)/5=0.1386$. Then [tex]1+ r= e^{0.1386}= 1.1486983550[tex] and so $r= 0.148693550$.