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Exponential Functions

Exponential Functions 1.4 - Solution

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To find the balance we will use the compound interest formula. In this formula, yy is the amount in the account after t\textcolor{darkorange}{t} years, P{\color{#0000FF}{P}} is the principal amount invested, r{\color{#FF0000}{r}} is the annual interest rate, and n{\color{#009600}{n}} is the number of compounding periods each year. y=P(1+rn)nt\begin{gathered} y={\color{#0000FF}{P}}\left(1+\dfrac{{\color{#FF0000}{r}}}{{\color{#009600}{n}}}\right)^{{\color{#009600}{n}}\textcolor{darkorange}{t}} \end{gathered} In our case, we know that t=7,\textcolor{darkorange}{t}=\textcolor{darkorange}{7}, P=700,{\color{#0000FF}{P}}={\color{#0000FF}{700}}, and r=4.3%=0.043.{\color{#FF0000}{r}}={\color{#FF0000}{4.3\%}}={\color{#FF0000}{0.043}}. Since there are 1212 months in one year, the number of compounding periods is n=12.{\color{#009600}{n}}={\color{#009600}{12}}. Let's substitute these values into the compound interest formula and calculate the balance after 7\textcolor{darkorange}{7} years in our deposit.
y=P(1+rn)nty={\color{#0000FF}{P}}\left(1+\dfrac{{\color{#FF0000}{r}}}{{\color{#009600}{n}}}\right)^{{\color{#009600}{n}}\textcolor{darkorange}{t}}
y=700(1+0.04312)127y={\color{#0000FF}{700}}\left(1+\dfrac{{\color{#FF0000}{0.043}}}{{\color{#009600}{12}}}\right)^{{\color{#009600}{12}}\cdot \textcolor{darkorange}{7}}
y945.34y\approx 945.34
We found that, after 77 years, the balance will be around $945.34.\$ 945.34.