So if I have A=xe^rt for a continuously compounding interest equation, then according to the calculator that is also equal to xLn(A)=rt, but when I solved for it using real numbers, I didn't get the correct number.

 

 

The known info is the initial amount is 750, and after 7 and 3/4 years, that amount doubles,

I have 1500=750e^r(7+3/4)

yet when I solved for it

 

750ln(1500)=r(7+3/4) ---> 5484.915=r(7+3/4), then /(7+3/4) and I got 707.731=r, and since I'm suppose to find out how much I will have after a specific amount of time as well, and I can't really go backwards in time, this doesn't work.

 

Basically, I am trying to solve for the annual rate, but it's not working for some reason.

Edited October 24, 201114 yr by questionposter

So if I have A=xe^rt for a continuously compounding interest equation, then according to the calculator that is also equal to xLn(A)=rt, but when I solved for it using real numbers, I didn't get the correct number.

 

 

The known info is the initial amount is 750, and after 7 and 3/4 years, that amount doubles,

I have 1500=750e^r(7+3/4)

yet when I solved for it

 

750ln(1500)=r(7+3/4) ---> 5484.915=r(7+3/4), then /(7+3/4) and I got 707.731=r, and since I'm suppose to find out how much I will have after a specific amount of time as well, and I can't really go backwards in time, this doesn't work.

 

Basically, I am trying to solve for the annual rate, but it's not working for some reason.

 

Couple of readability tips to make people enjoy reading your questions more (and just be more likely to answer).

 

1. Brackets please.e^rt is somewhat ambiguous as to whether you mean e^(rt) or (e^r)t.

2. (Not as necessary) Read up on the latex tutorial in the maths section if you're posting a lot of maths. It makes it a bit easier to read. That equation would be something along the lines of:

[math]A=xe^{(rt)}[/math]

Which looks like: [math]A=xe^{(rt)}[/math]

 

(You can also click on the images to pull up the latex that produced them)

 

If you use latex then the brackets aren't necessary (as all the indeces will be superscript), but I left them in to demonstrate what I meant.

 

Doing the rearrangement:

 

[math]A=xe^{(rt)}[/math]

[math]\implies \ln{(A)}=\ln{\left( x \times e^{(rt)}\right)}[/math]

[math]\implies \ln{(A)}=\ln{(x)} + (rt)[/math]

Edited October 24, 201114 yr by Schrödinger's hat

Couple of readability tips to make people enjoy reading your questions more (and just be more likely to answer).

 

1. Brackets please.e^rt is somewhat ambiguous as to whether you mean e^(rt) or (e^r)t.

2. (Not as necessary) Read up on the latex tutorial in the maths section if you're posting a lot of maths. It makes it a bit easier to read. That equation would be something along the lines of:

[math]A=xe^{(rt)}[/math]

Which looks like: [math]A=xe^{(rt)}[/math]

 

(You can also click on the images to pull up the latex that produced them)

 

If you use latex then the brackets aren't necessary (as all the indeces will be superscript), but I left them in to demonstrate what I meant.

 

Doing the rearrangement:

 

[math]A=xe^{(rt)}[/math]

[math]\implies \ln{(A)}=\ln{\left( x \times e^{(rt)}\right)}[/math]

[math]\implies \ln{(A)}=\ln{(x)} + (rt)[/math]

 

Perhaps I did it wrong, but when I solved I got about .038, and when I looked on the table for the equation 750e^(.038x), it did not equal 1500 after 7 and 3/4 years.

 

When you said

 

[math]\implies \ln{(A)}=\ln{(x)} + (rt)[/math]

 

did you mean Ln(x + rt) or did you mean (ln(x))+rt?

Edited October 24, 201114 yr by questionposter

The rule is as follows:

 

[math]\text{ln}(a \times b) = \text{ln}(a) + \text{ln}(b)[/math]

 

We also know that:

 

[math]\text{ln}(a^x)=x \, \text{ln} (a)[/math]

 

and

 

[math]\text{ln}(e)=1[/math]

 

So we can conclude that:

 

[math]\text{ln}(x \times e^{r\, t})=\text{ln}(x)+\text{ln}(e^{r\, t})=\text{ln}(x)+r\, t \, \text{ln}(e)=\text{ln}(x)+r\, t[/math]

Edited October 24, 201114 yr by Daedalus

The rule is as follows:

 

[math]\text{ln}(a \times b) = \text{ln}(a) + \text{ln}(b)[/math]

 

We also know that:

 

[math]\text{ln}(a^x)=x \, \text{ln} (a)[/math]

 

and

 

[math]\text{ln}(e)=1[/math]

 

So we can conclude that:

 

[math]\text{ln}(x \times e^{r\, t})=\text{ln}(x)+\text{ln}(e^{r\, t})=\text{ln}(x)+r\, t \, \text{ln}(e)=\text{ln}(x)+r\, t[/math]

 

Thanks, that seemed to be what was missing.

You are welcome : )

 

 

Superb Daedalus !

 

Really Nice easy to clear everything.