Determining monthly interest rate after money has already accumulated(growth rate)

[perseverer]

Hi Everyone.
I wanted to know if anyone could help me determine at what rate my savings grew from $68 to 470 dollars over an 11 month period. I found that the compound interest rate formula is P=C(1+r/n) ^nt but sadly I don't know how to solve for r. I would like to know at what percent per month my money grew.
P=future value
C=Initial Deposit
r=interest rate
n=number of times per year the money is compounded
t= the number of years.
Any help would be very much appreciated.
Thank you.

 

[tkhunny]

You will need a logarithm. If you have no familiarity with the logarithm, you cannot do it or understand it.

Well, what say you? Logarithms or no?

 

[perseverer]

Log please

You will need a logarithm. If you have no familiarity with the logarithm, you cannot do it or understand it.

Well, what say you? Logarithms or no?

Thank you for the response tkhunny.
If a logarithm is what I need then by George a logarithm is indeed what I want. Have you any logarithms to spare? ;-)
Is it not possible to solve for r?
Would I need a different log for a 15 month period?
How does one make or determine a log?
Thanks again for your time and knowledge.

 

[tkhunny]

One free one, but promise me you'll find a text book and learn what these guys are!

P=C(1+r/n) ^nt

Divide by C

P/C = (1+ r/n)^(n*t)

LOGARITHM TIME

log(P/C) = log[(1+ r/n)^(n*t)] = (n*t)*log(1 + r/n)

Divide by n*t

\(\displaystyle \frac{log(P/C)}{n*t} = log(1 + r/n)\)

LOGARITHM TIME Again, in reverse

\(\displaystyle \left(\frac{P}{C}\right)^\frac{1}{n\cdot t} = 1 + r/n\)

Subtract 1 and multiply by n

\(\displaystyle \left[\left(\frac{P}{C}\right)^\frac{1}{n\cdot t} - 1\right]\cdot n = r\)

I didn't say it would be pretty.

 

[perseverer]

Thank you for your time, knowledge, and kindness

Now I see that there are really useful applications for logs. I just joined this great site and so I will search for log study materials. Thank you once again

 

[perseverer]

One more question, in my case then is this right?

[(470/68)^1/11 -1] Yes? Can I ignore the 't' or is it 11/12? Sorry for my ignorance. Thanks again

 

[perseverer]

Thank you Denis

Thanks Denis. Now I have a question about the actual description of what has occurred with my money. Namely the sum has increased almost 7 times which is x% higher but we say that the annual interest rate is 230%? Sorry for my math illiteracy. The more questions I ask the more I expose my ignorance.
If i could communicate with numbers the way you can I would probably be a man of few words too.

 

[tkhunny]

You can do it without logs

Actually, no. It is the same. Just because you don't SAY "log", doesn't mean you didn't use them.

If you use this definition, Logarithm = Exponent, it's still a log.

 

[perseverer]

Thank you so much

This is a great forum. After I learn more I hope to conribute as well. Thank you.