Answer:
i) See proof below.
ii) 5 years
Step-by-step explanation:
To solve this problem, we can use the formula for annual compound interest:
[tex]\boxed{\begin{minipage}{7 cm}\underline{Annual Compound Interest Formula}\\\\$ A=P\left(1+r\right)^{t}$\\\\where:\\\\ \phantom{ww}$\bullet$ $A =$ final amount \\ \phantom{ww}$\bullet$ $P =$ principal amount \\ \phantom{ww}$\bullet$ $r =$ interest rate (in decimal form) \\ \phantom{ww}$\bullet$ $t =$ time (in years) \\ \end{minipage}}[/tex]
Given values:
- A = $2036
- P = $2000
- t = 1 year
Substitute the values into the formula and solve the equation for r:
[tex]2036=2000(1+r)^1[/tex]
[tex]2036=2000(1+r)[/tex]
[tex]\dfrac{2036}{2000}=1+r[/tex]
[tex]1.018=1+r[/tex]
[tex]r=1.018-1[/tex]
[tex]r=0.018[/tex]
Multiply by 100 to convert to a percentage:
[tex]r = 0.018 \times 100=1.8\%[/tex]
Therefore, the interest rate per year, K, is 1.8%.
To find the number of complete years before Nicole has at least $2,150 in the account, substitute A = 2150, P = 2000, and r = 0.018 into the formula, and solve for t.
[tex]2150=2000(1+0.018)^t[/tex]
[tex]2150=2000(1.018)^t[/tex]
[tex]\dfrac{2150}{2000}=(1.018)^t[/tex]
[tex]1.075=(1.018)^t[/tex]
Take natural logs of both sides of the equation:
[tex]\ln 1.075= \ln (1.018)^t[/tex]
[tex]\textsf{Apply the log power law:} \quad \ln x^n=n \ln x[/tex]
[tex]\ln 1.075= t \ln 1.018[/tex]
Divide both sides of the equation by ln(1.018):
[tex]\dfrac{\ln 1.075}{\ln 1.018}= \dfrac{t \ln 1.018}{\ln 1.018}[/tex]
[tex]t=4.05386734...[/tex]
As we need to find the number of complete years before Nicole has at least $2,150 in the account, we need to round up the found value of t to the nearest complete year.
From our calculations, t ≈ 4.05 years, so the balance of the account reaches $2,150 after 4 years and during the 5th year. (After 4 years, the account balance is $2,147.93).
Therefore, the number of complete years before Nicole has at least $2,150 in the account is 5 years.
