## jcdavi34 one year ago Allie currently has an account balance of $2,265.96. She opened the account 13 years ago with a deposit of$1,227.43. If the interest compounds daily, what is the interest rate on the account?

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1. Hunus

The formula for compounded interest is $A=P(1+\frac{r}{n})^{nt}$ Where n is the number of times per year it is compounded So the equation you will need to use is $A=P(1+\frac{r}{365})^{365t}$

2. jcdavi34

i got 1.9%

3. Hunus

I got something different. Will you show me your steps?

4. jcdavi34

i got 4.7%

5. jcdavi34

2265.96(1+r/365)365*12

6. jcdavi34

...........

7. Hunus

Okay is it 12 or 13 years?

8. jcdavi34

no

9. jcdavi34

intrest rate

10. Hunus

no? In your question you put 13 in your step you put 12

11. Hunus

I know. But I need to know if t = 12 or t = 13

12. jcdavi34

whats r

13. Hunus

$2265.96=1227.43(1+\frac{r}{365})^{365*13}$ $\frac{2265.96}{1227.43}=(1+\frac{r}{365})^{365*13}$ $\log(1.8461)=\log((1+\frac{r}{365})^{365*13})$ $\log(1.8461)=365*13*\log((1+\frac{r}{365}))$ $\frac{\log(1.8461)}{365*13}=1.000129=\log(1+\frac{r}{365})$ $10^{1.000129}=1+\frac{r}{365}$ $10^{1.000129}-1=0.000129=\frac{r}{365}$ $r=0000129*365=0.0471$