Question

Suppose that $1000 is deposited into an account where the interest is compounded annually. This situation can be modeled by the function, f(t) = 1000(1.06), where f(t) represents the value (in dollars) of the account at t years after depositing the $1000. 1. In how many years will the money in the account double? 2. According to this model, how much money will the account have in 10 years? 3. According to this model, what is interest rate being earned per year.

Answer 1

$f(t)=1000(1.06)^t$

As the initial deposited is $1000

1. In how many years will the money in the acount be 2000

$\begin{gathered} f(t)=2000 \\ \\ 2000=1000(1.06)^t \end{gathered}$

To solve for t:

-Divide both sides of the equation into 1000

$\begin{gathered} \frac{2000}{1000}=\frac{1000(1.06)^t}{1000} \\ \\ 2=(1.06)^t \end{gathered}$

- As:

$\begin{gathered} a^b=c \\ \text{Log}_ac=b \end{gathered}$$\begin{gathered} (1.06)^t=2 \\ \log _{1.06}2=t \end{gathered}$

-As:

$\log _ac=\frac{\log c}{\log a}$$\log _{1.06}2=\frac{\log 2}{\log 1.06}=11.89$Then, after 11.89 or approximately 12 years the money will be 2000.---------------------------

To find the amount of money after 10 years you substitute the t in the equation for 10 and evaluate the function:

$\begin{gathered} f(t)=1000(1.06)^t \\ f(10)=1000(1.06)^{10} \\ f(10)=1000(1.7908)=1790.8 \end{gathered}$Then, after 10 years the amount of money in the account is $1790.8-----------------------

The general equation for a compounded interest is:

$f(t)=C(1+i)^t$

Where C is the initial deposit

i is the interest (in decimal)

As you have:

$1+i=1.06$

the interest in decimal will be:

$\begin{gathered} 1.06-1=i \\ 0.06=i \end{gathered}$Then, the interest rate is: 6%$\begin{gathered} i\cdot100 \\ =0.06\cdot100 \\ =6 \end{gathered}$