the formula a = p(1 + r/n)^(nt) describes the accumulated value, a, of a sum of money, p, the principal…

the formula a = p(1 + 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| $10,500 | 2 | 4.75% | $16,000 |? |\nt ≈ years\n(do not round until the final answer. then round to one decimal place as needed.)

the formula a = p(1 + 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| $10,500 | 2 | 4.75% | $16,000 |? |\nt ≈ years\n(do not round until the final answer. then round to one decimal place as needed.)

Answer

Explanation:

Step1: Identify values

$P = 10500$, $n = 2$, $r=0.0475$, $A = 16000$

Step2: Substitute into formula

$16000=10500\left(1+\frac{0.0475}{2}\right)^{2t}$

Step3: Simplify the equation

$\frac{16000}{10500}=\left(1 + 0.02375\right)^{2t}$ $\frac{32}{21}=(1.02375)^{2t}$

Step4: Take the natural - logarithm of both sides

$\ln\left(\frac{32}{21}\right)=\ln\left((1.02375)^{2t}\right)$ Using the property $\ln(a^b)=b\ln(a)$, we get $\ln\left(\frac{32}{21}\right)=2t\ln(1.02375)$

Step5: Solve for $t$

$t=\frac{\ln\left(\frac{32}{21}\right)}{2\ln(1.02375)}$ $t=\frac{\ln(32)-\ln(21)}{2\ln(1.02375)}$ $\ln(32)\approx3.4657$, $\ln(21)\approx3.0445$, $\ln(1.02375)\approx0.0235$ $t=\frac{3.4657 - 3.0445}{2\times0.0235}=\frac{0.4212}{0.047}\approx8.96$

Answer:

$9.0$