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

Write the exponential function for the following table andcomplete the table.Value-example-1
User JustJeff
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1 Answer

25 votes
25 votes

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}/\text{ equation \lparen1\rparen;} \\ (1800)/(1200)=(ab^(10))/(ab^5) \\ 1.5=b^5 \\ b=1.5^{(1)/(5)} \\ SInce\text{ the initial is \$800,} \\ a=800 \\ \\ \\ Hence,\text{ the function is:} \\ y=ab^x \\ y=800(1.5)^{(1)/(5)* x} \\ \\ y=800(1.5)^{(x)/(5)} \end{gathered}

Therefore, the exponential function is:


y=800(1.5)^{(x)/(5)}

To complete the table,

when x = 20 years;


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

when x = 25 years;


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

when x = 30 years;


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

when x = 35 years;


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

Write the exponential function for the following table andcomplete the table.Value-example-1
User Terris
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2.9k points