the formula a = p(1 + \\frac{r}{n})^{nt} describes the accumulated value, a, of a sum of money, p, the…

the formula a = p(1 + \\frac{r}{n})^{nt} describes the accumulated value, a, of a sum of money, p, the principal, after t years at annual percentage rate r (in decimal form) compounded n times a year. complete the table for a savings account subject to n compounding periods per year.\n| amount invested | number of compounding periods | annual interest rate | accumulated amount | time t in years |\n| $14,000 | 2 | 5.25% | $20,000 |? |\n\nt ≈ □ years\n(do not round until the final answer. then round to one decimal place as needed.)

the formula a = p(1 + \\frac{r}{n})^{nt} describes the accumulated value, a, of a sum of money, p, the principal, after t years at annual percentage rate r (in decimal form) compounded n times a year. complete the table for a savings account subject to n compounding periods per year.\n| amount invested | number of compounding periods | annual interest rate | accumulated amount | time t in years |\n| $14,000 | 2 | 5.25% | $20,000 |? |\n\nt ≈ □ years\n(do not round until the final answer. then round to one decimal place as needed.)

Answer

Explanation:

Step1: Identify values

$P = 14000$, $n = 2$, $r=0.0525$, $A = 20000$

Step2: Substitute into formula

The compound - interest formula is $A = P(1+\frac{r}{n})^{nt}$. Substituting the values, we get $20000=14000(1 +\frac{0.0525}{2})^{2t}$.

Step3: Simplify the equation

First, divide both sides by 14000: $\frac{20000}{14000}=(1 + 0.02625)^{2t}$, so $\frac{10}{7}=(1.02625)^{2t}$.

Step4: Take the natural logarithm of both sides

$\ln(\frac{10}{7})=\ln((1.02625)^{2t})$. Using the property $\ln(a^b)=b\ln(a)$, we have $\ln(\frac{10}{7}) = 2t\ln(1.02625)$.

Step5: Solve for t

$t=\frac{\ln(\frac{10}{7})}{2\ln(1.02625)}$. $\ln(\frac{10}{7})\approx\ln(10)-\ln(7)\approx2.3026 - 1.9459=0.3567$, and $\ln(1.02625)\approx0.0259$. $2\ln(1.02625)\approx0.0518$. $t=\frac{0.3567}{0.0518}\approx6.9$.

Answer:

$6.9$