find the rate of compound interest

[Ginandtonic]

Hello. I've been given some homework to do at home. I need to know how to find out the % rate of compound interest when I have the original amount of money to be saved and I also have the final amount saved after the interest is added, and I also know the number of years it was saved for. Can any one help me please? Thank you.

 

[Romsek]

Do you know what the formula is for the future value of an account given the time, rate, and original value of the account is?
It's a pretty standard formula you are no doubt expected to know.

 

[Ginandtonic]

I use this to work out the final amount of money, but don't know how to use it to work out the interest rate if I already have the final amount of money: amount saved (1+interest rate/100) to the power of number of years

 

[pka]

I use this to work out the final amount of money, but don't know how to use it to work out the interest rate if I already have the final amount of money: amount saved (1+interest rate/100) to the power of number of years

The best way for all us is if you will post the exact question in full. That way we can see what is required to help you solve it.

 

[Romsek]

I use this to work out the final amount of money, but don't know how to use it to work out the interest rate if I already have the final amount of money: amount saved (1+interest rate/100) to the power of number of years

expressing that mathematically would have been nice but you have the gist of it
[MATH] f = p\left(1+\dfrac{r}{100}\right)^y\\ \text{where $r$ is the rate in percentage and $y$ is the number of years}\\ \text{Now you say you are given the future value $f$, and present value $p$, and the number of years $y$}\\ \text{So let's solve for $r$ as a percentage}\\ \log(f) = \log\left(p \left(1+\dfrac{r}{100}\right)^y\right)\\ \log(f) = \log(p) + y\log\left(1+\dfrac{r}{100}\right)\\ \dfrac{\log(f)-\log(p)}{y} = \log\left(1+\dfrac{r}{100}\right)\\ 10^{\frac{\log(f)-\log(p)}{y}} = 1+\dfrac{r}{100}\\ r = 100 \left(10^{\frac{\log(f)-\log(p)}{y}}-1\right) = 100 \left(10^{\frac{\log(f/p)}{y}}-1\right) [/MATH]

 

[Ginandtonic]

expressing that mathematically would have been nice but you have the gist of it
[MATH] f = p\left(1+\dfrac{r}{100}\right)^y\\ \text{where $r$ is the rate in percentage and $y$ is the number of years}\\ \text{Now you say you are given the future value $f$, and present value $p$, and the number of years $y$}\\ \text{So let's solve for $r$ as a percentage}\\ \log(f) = \log\left(p \left(1+\dfrac{r}{100}\right)^y\right)\\ \log(f) = \log(p) + y\log\left(1+\dfrac{r}{100}\right)\\ \dfrac{\log(f)-\log(p)}{y} = \log\left(1+\dfrac{r}{100}\right)\\ 10^{\frac{\log(f)-\log(p)}{y}} = 1+\dfrac{r}{100}\\ r = 100 \left(10^{\frac{\log(f)-\log(p)}{y}}-1\right) = 100 \left(10^{\frac{\log(f/p)}{y}}-1\right) [/MATH]

Thank you - but what does log mean please?

 

[Ginandtonic]

The original problem was: Find the interest rate if £12800 has a future value of £18605 in 4 years. Give your answer to 1 d.p. I'm 13 and don't think I'm very good at maths so if anyone could make it step by step that would be really good thank you.

 

[Oiler]

The original problem was: Find the interest rate if £12800 has a future value of £18605 in 4 years. Give your answer to 1 d.p. I'm 13 and don't think I'm very good at maths so if anyone could make it step by step that would be really good thank you.

A step by step and bottom up formulation of the solvable equation might involve the logical deduction that compound growth in this context involves the recalculation of the annual interest based upon the previous year's sum (as opposed to a constant, unchanging simple interest) such that:
[MATH]12800\cdot r=12800r[/MATH] [where r is the rate you want to calculate] and [MATH]12800r[/MATH] gives the sum of money at the end of one year. We proceed similarly three more times
[MATH]12800r\cdot r=12800r^2[/MATH] where [MATH]12800r^2[/MATH] denotes the sum of money at the end of two years
[MATH]12800r^2\cdot r=12800r^3[/MATH] ...(you get the memo, this is the sum at the end of three years)
finally, we reach an expression for the sum of money at the end of the fourth year:
[MATH]12800r^3 \cdot r=12800r^3[/MATH] and the full problem you posted happens to also tell us the final amount of money produced with the interest at the end of 4 years so this must mean that:
[MATH]12800r^4=18605[/MATH]Can you take it from here and solve for the interest rate, [MATH]r[/MATH]? ; )

 

[Ginandtonic]

Ok, thank you. So should I then work out a root?