I need help finding the equations for two problems.
Salary of $\$100,000$ increasing by $\$10,000$ each year for $5$ years.
Salary of $\$75,000$ increasing by $30\%$ each year for $5$ years.
I know that one will be an exponential equation and one will be a linear equation. I also know that for the salary increasing by $\$10,000$ each year the sums are:
Year 1: 100,000 2: 110,000 3: 120,000 4: 130,000 5: 140,000
and for the salary increasing by 30% each year the sums are:
year 1: 75,000 2: 97,500 3: 126,750 4: 164,775 5: 240,825
I believe the salary increasing by 10,000 each year might be: 100,000 + (100,000*.10t), because it increases by 10% of the initial 100,000 each year (t), but I'm not sure.
Thank you!
So, in the first case, you have \begin{equation} y(t)=100,000+(10,000\cdot t) \end{equation} The variable $t$ is multypling a constant and the function $y(t)$ is a polynomial, which means that $y(t)$ is a linear function, and is linear in $t$.
For the second question, consider that: \begin{equation} y_2=y_1+0.3\cdot y_1= 1.3 \cdot y_1 \end{equation} \begin{equation} y_3= 1.3 \cdot y_2 \end{equation}
\begin{equation} \vdots \end{equation}
\begin{equation} y_t= 1.3 \cdot y_{t-1} \end{equation} Which, substituting in a single expression, gives: \begin{equation} y_t=(1.3)^t \cdot y_1 \end{equation} that is if you take as $t=0$ as your starting point at $y_1=75,000$. If your starting point is $t=1$, then you simply have: \begin{equation} y(t)=(1.3)^{(t-1)} \cdot y_1 \end{equation} This equation has the variable as the exponent of a constant, hence the linearity is violated and the function is said to be exponential in $t$. And by the way check yours calculations, the amounts per year are: \begin{equation} y_1=$75,000\\ y_2=$97,500\\ y_3=$126,750\\ y_4=$164,775\\ y_5=$214,2075\\ \end{equation}
These are the plots: Plot 1 Plot 2