Question

Find the exponential function that models the data in the table below.

Answer 1

An exponential function is defined by

`f(x)=Ae^(Bx+C)`

Where x occurs as an exponent.

The exponential curve depends on the exponential function and it depends on the value of the x. Where a>0 and a is not equal to 1. x is any real number.

If we graph the data table we can see that:

From the general equation we take the natural logarithm to both sides to be able to solve for A, B and C easily. With the points in the table we will make the different equations, as we have 3 unknowns we must have at least 3 equations

We can see graphically that there is no phase shift in the function, which is why C = 0

`\begin{gathered} \ln (f(x))=Ln(A\cdot e^(Bx)) \\ \ln (f(x))=Ln(A)+Ln(e^(Bx)) \\ \ln (f(x))=Ln(A)+(Bx) \end{gathered}`

We have the values of x and F(x) from the tables, we are going to replace them to find the equations

`\begin{gathered} Ln(2)=Ln(A)+B(0)+0 \\ 0.69=Ln(A)\to(1) \\ e^(0.69)=A \\ A=1.99 \end{gathered}`
`\begin{gathered} Ln(6)=Ln(A)+B(1) \\ 1.79=Ln(A)+B\to(2) \end{gathered}`
`\begin{gathered} Ln(18)=Ln(A)+B(2)+C \\ 2.89=Ln(A)+2B+C\to(3) \end{gathered}`
`\begin{gathered} Ln(54)=Ln(A)+B(3) \\ 3.99=Ln(A)+3B\to(4) \end{gathered}`

Now we have the 4 equations and we have 2 unknowns we will solve between them to find the values of A, B and C

`\begin{gathered} Ln(A)=0.69\to(1) \\ 1.79=0.69+B\to(1)\text{ in (2)} \\ 1.1=B \end{gathered}`

Then our equation would be:

`\begin{gathered} f(x)=1.99\cdot e^(1.1\cdot x) \\ \end{gathered}`
`y=1.99\cdot e^(1.1\cdot x)`

You can round the function as follows:

`y=2e^(1.1x)`