madeline invested $960 in an account paying an interest rate of 3.5% compounded daily. assuming no deposits…

Explanation:

Step1: Recall compound - interest formula

The compound - interest formula when compounded daily is $A = P(1+\frac{r}{n})^{nt}$, where $A$ is the final amount, $P$ is the principal amount, $r$ is the annual interest rate (in decimal), $n$ is the number of times compounded per year, and $t$ is the number of years. Here, $P = 960$, $A = 1250$, $r=0.035$ (since $3.5%=0.035$), and $n = 365$. So, $1250=960(1 +\frac{0.035}{365})^{365t}$.

Step2: Divide both sides by 960

$\frac{1250}{960}=(1+\frac{0.035}{365})^{365t}$. $\frac{125}{96}=(1+\frac{0.035}{365})^{365t}$. $1+\frac{0.035}{365}=1+\frac{35}{365000}=1+\frac{7}{73000}\approx1.00009589$. So, $\frac{125}{96}\approx(1.00009589)^{365t}$.

Step3: Take the natural logarithm of both sides

$\ln(\frac{125}{96})=\ln((1.00009589)^{365t})$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln(\frac{125}{96}) = 365t\ln(1.00009589)$. $\ln(\frac{125}{96})=\ln(125)-\ln(96)\approx4.8283137 - 4.564347\approx0.263967$. $\ln(1.00009589)\approx0.00009589$.

Step4: Solve for $t$

$t=\frac{\ln(\frac{125}{96})}{365\times\ln(1.00009589)}$. $365\times\ln(1.00009589)\approx365\times0.00009589 = 0.03499985$. $t=\frac{0.263967}{0.03499985}\approx7.54$.

Answer:

8 years