Question

Write the exponential function for the following table andcomplete the table.Value ofTimeInvestmentInitial$8005 yr$120010 yr$180015 yr$270020 yr25 y30 yr35 yr

Answer 1

Solution:

Given:

Using the exponential function formula,

$y=ab^x$

From the table, pick two points; (5,1200) and (10,1800)

$\begin{gathered} 1200=ab^5...............(1) \\ 1800=ab^{10}.....................(2) \\ Equation\text{ \lparen2\rparen}\div\text{ equation \lparen1\rparen;} \\ \frac{1800}{1200}=\frac{ab^{10}}{ab^5} \\ 1.5=b^5 \\ b=1.5^{\frac{1}{5}} \\ SInce\text{ the initial is \ 800,} \\ a=800 \\ \\ \\ Hence,\text{ the function is:} \\ y=ab^x \\ y=800(1.5)^{\frac{1}{5}\times x} \\ \\ y=800(1.5)^{\frac{x}{5}} \end{gathered}$

Therefore, the exponential function is:

$y=800(1.5)^{\frac{x}{5}}$

To complete the table,

when x = 20 years;

$\begin{gathered} y=800(1.5)^{\frac{x}{5}} \\ y=800(1.5)^{\frac{20}{5}} \\ y=800(1.5^4) \\ y=\text{ \ }4050 \end{gathered}$

when x = 25 years;

$\begin{gathered} y=800(1.5)^{\frac{x}{5}} \\ y=800(1.5)^{\frac{25}{5}} \\ y=800(1.5^5) \\ y=\text{ \ }6075 \end{gathered}$

when x = 30 years;

$\begin{gathered} y=800(1.5)^{\frac{x}{5}} \\ y=800(1.5)^{\frac{30}{5}} \\ y=800(1.5^6) \\ y=\text{ \ }9112.5 \end{gathered}$

when x = 35 years;

$\begin{gathered} y=800(1.5)^{\frac{x}{5}} \\ y=800(1.5)^{\frac{35}{5}} \\ y=800(1.5^7) \\ y=\text{ \ }13668.75 \end{gathered}$