Eric made two investments: $\newline$ - Investment $\mathrm{Q}$ has a value of $\$ 500$ at the end of the first year and increases by $\$ 45$ per year. $\newline$ - Investment $\mathrm{R}$ has a value of $\$ 400$ at the end of the first year and increases by $10 \%$ per year. $\newline$ Eric checks the value of his investments once a year, at the end of the year. $\newline$ What is the first year in which Eric sees that investment $R$ 's value exceeded investment Q's value?

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Algebra 2

Interpret the exponential function word problem

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Q. Eric made two investments:

\newline

- Investment

\mathrm{Q}

has a value of

\$ 500

at the end of the first year and increases by

\$ 45

per year.

\newline

- Investment

\mathrm{R}

has a value of

\$ 400

at the end of the first year and increases by

10 \%

per year.

\newline

Eric checks the value of his investments once a year, at the end of the year.

\newline

What is the first year in which Eric sees that investment

R

's value exceeded investment Q's value?

Define $Q(n)$ : Step $1$ : Define the value of investment $Q$ after $n$ years. $\newline$ Investment $Q$ starts at $\$500$ and increases by $\$45$ each year. So, the value of investment $Q$ after $n$ years can be represented as: $\newline$ $Q(n) = 500 + 45n$
Define $R(n)$ : Step $2$ : Define the value of investment $R$ after $n$ years. $\newline$ Investment $R$ starts at $\$400$ and increases by $10\%$ each year. The value of investment $R$ after $n$ years can be represented as: $\newline$ $R(n) = 400 \times (1 + 0.10)^n$
Determine year when $R(n)$ exceeds $Q(n)$ : Step $3$ : Determine the year when $R(n)$ exceeds $Q(n)$ . $\newline$ We need to find the smallest integer $n$ for which $R(n) > Q(n)$ . This means we are looking for the smallest $n$ such that: $\newline$ $400 \times (1.10)^n > 500 + 45n$
Solve inequality: Step $4$ : Solve the inequality by trial and error or algebraically. $\newline$ Since the inequality involves an exponential term and a linear term, it's not straightforward to solve algebraically. We will use trial and error, starting with $n = 1$ and increasing $n$ until $R(n)$ exceeds $Q(n)$ .
Calculate $Q(n)$ and $R(n)$ for $n=1$ : Step $5$ : Calculate the value of $Q(n)$ and $R(n)$ for successive years. $\newline$ For $n = 1$ : $\newline$ $Q(1) = 500 + 45(1) = 545$ $\newline$ $R(1) = 400 \times (1.10)^1 = 440$ $\newline$ $R(1)$ is not greater than $Q(1)$ .
Increment $n$ and calculate: Step $6$ : Increment $n$ and calculate again. $\newline$ For $n = 2$ : $\newline$ $Q(2) = 500 + 45(2) = 590$ $\newline$ $R(2) = 400 \times (1.10)^2 = 484$ $\newline$ $R(2)$ is not greater than $Q(2)$ .
Continue incrementing $n$ : Step $7$ : Continue incrementing $n$ . $\newline$ For $n = 3$ : $\newline$ $Q(3) = 500 + 45(3) = 635$ $\newline$ $R(3) = 400 \times (1.10)^3 = 532.4$ $\newline$ $R(3)$ is not greater than $Q(3)$ .
Continue incrementing $n$ : Step $8$ : Continue incrementing $n$ . $\newline$ For $n = 4$ : $\newline$ $Q(4) = 500 + 45(4) = 680$ $\newline$ $R(4) = 400 \times (1.10)^4 = 585.64$ $\newline$ $R(4)$ is not greater than $Q(4)$ .
Continue incrementing $n$ : Step $9$ : Continue incrementing $n$ . $\newline$ For $n = 5$ : $\newline$ $Q(5) = 500 + 45(5) = 725$ $\newline$ $R(5) = 400 \times (1.10)^5 = 644.204$ $\newline$ $R(5)$ is not greater than $Q(5)$ .
Continue incrementing $n$ : Step $10$ : Continue incrementing $n$ . $\newline$ For $n = 6$ : $\newline$ $Q(6) = 500 + 45(6) = 770$ $\newline$ $R(6) = 400 \times (1.10)^6 = 708.624$ $\newline$ $R(6)$ is not greater than $Q(6)$ .
Continue incrementing $n$ : Step $11$ : Continue incrementing $n$ . $\newline$ For $n = 7$ : $\newline$ $Q(7) = 500 + 45(7) = 815$ $\newline$ $R(7) = 400 \times (1.10)^7 = 779.4864$ $\newline$ $R(7)$ is not greater than $Q(7)$ .
Continue incrementing $n$ : Step $12$ : Continue incrementing $n$ . $\newline$ For $n = 8$ : $\newline$ $Q(8) = 500 + 45(8) = 860$ $\newline$ $R(8) = 400 \times (1.10)^8 = 857.43504$ $\newline$ $R(8)$ is not greater than $Q(8)$ .
Continue incrementing $n$ : Step $13$ : Continue incrementing $n$ . $\newline$ For $n = 9$ : $\newline$ $Q(9) = 500 + 45(9) = 905$ $\newline$ $R(9) = 400 \times (1.10)^9 = 943.178544$ $\newline$ $R(9)$ is greater than $Q(9)$ . Therefore, the first year in which investment R's value exceeded investment Q's value is after $9$ years.