Respuesta :
Answer:
6.33 years
Explanation:
The formula for investment at compound interest is given below::
[tex]A(t)=P\left(1+\frac{r}{k}\right)^{tk}\text{ where }\begin{cases}P=\text{Principal Invested} \\ r=\text{Interest Rate} \\ k=\text{Number of compounding periods}\end{cases}[/tex]From the statement of the problem:
• The initial investment, P = $2500
,• Annual Interest Rate, r = 7.5% = 0.075
,• Compounding Period (Quarterly), k = 4
,• Amount after t years, A(t) = $4000
,• Time, t = ?
Substitute these values into the compound interest formula above:
[tex]4000=2500\left(1+\frac{0.075}{4}\right)^{4t}[/tex]We then solve the equation for the value of t.
[tex]\begin{gathered} \begin{equation*} 4000=2500\left(1+\frac{0.075}{4}\right)^{4t} \end{equation*} \\ \text{ Divide both sides by 2500} \\ \frac{4000}{2500}=\left(1+0.01875\right)^{4t} \\ 1.6=\left(1.01875\right)^{4t} \\ \text{ Take the log of both sides} \\ \log(1.6)=\log(1.01875)^{4t} \\ \text{ By the power law of logs, }\log a^n=n\log a \\ \log(1.6)=4t\log(1.01875) \\ \text{ Divide both sides by 4}\log(1.01875) \\ \frac{\operatorname{\log}(1.6)}{4\operatorname{\log}(1.01875)}=\frac{4t\operatorname{\log}(1.01875)}{4\operatorname{\log}(1.01875)} \\ t\approx6.33\text{ years} \end{gathered}[/tex]It will take approximately 6.33 years for a $2500 investment to grow to $4000.
