Finding the equation of an exponential function from a table

[smileyface497]

Given the table

t | h(t)
0.0 | 2.04
1.0 | 3.06
2.0 | 4.59
3.0 | 6.89
4.0 | 10.33
5.0 | 15.49

find the equation of the exponential function.

How would I go about doing this?

 

[Deleted member 4993]

Re: Finding the equation of an exponential function from a t

Given the table

t | h(t)
0.0 | 2.04
1.0 | 3.06
2.0 | 4.59
3.0 | 6.89
4.0 | 10.33
5.0 | 15.49

find the equation of the exponential function.

How would I go about doing this?

Assume:

$\displaystyle f(t) \ = \ A \ * \ e^{B*t}$

 

[soroban]

Re: Finding the equation of an exponential function from a t

Hello, smileyface497!

Given the table:

. . $\displaystyle \begin{array}{c|c} t & h(t) \\ \hline 0.0 & 2.04 \\ 1.0 & 3.06 \\ 2.0 & 4.59 \\ 3.0 & 6.89 \\ 4.0 & 10.33 \\ 5.0 & 15.49 \end{array}$

Find the equation of the exponential function.

$\displaystyle \text{The general exponential function would be: }\;h(t) \;=\;a\cdot b^{t} + c$


From the first four values of the table, we have:

. . $\displaystyle \begin{array}{cccccccc}h(0)\:=\:2.04 & a + c &=& 2.4 & [1] \\ h(1) \:=\: 3.06 & a\cdot b + c &=& 3.06 & [2] \\ h(2) \:=\:4.59 & a\cdot b^2 + c &=& 4.59 & [3] \\ h(3) \:=\:6.89 & a\cdot b^3 + c &=& 6.89 & [4] \end{array}$


$\displaystyle \begin{array}{cccccccccc}\text{Subtract [2] - [1]:} & ab - a &=& 1.02 & \Rightarrow & a(b-1) &=& 1.02 & [5] \\ \text{Subtract [3] - [2]:} & ab^2 - ab &=& 1.53 & \Rightarrow & ab(b-1) &=& 1.53 & [6] \\ \text{Subtract [4] - [3]:} & ab^3 - ab^2 &=& 2.30 & \Rightarrow & ab^2(b-1) &=& 2.30 & [7] \end{array}$


\(\displaystyle \begin{array}{cccccccccc} \text{Divide [6] by [5]:} & \dfrac{ab(b-1)}{a(b-1)} &=& \dfrac{1.53}{1.02} & \Rightarrow & b &=& 1.5\qquad \\ \\[-3mm] \text{Divide [7] by [6]:} & \dfrac{ab^2(b-1)}{ab(b-1)} &=& \dfrac{2.30}{1.53} & \Rightarrow & b &=& 1.495... \end{array} \;\; \Rightarrow\quad \boxed{b \;\approx\;1.5}\)


$\displaystyle \begin{array}{cccccccc}\text{Substitute into [2]:} & 1.5a + d &=& 3.06 & [8] \\ \text{Substitute into [3]:} & 2.25a + d &=& 4.59 & [9] \end{array}$

$\displaystyle \text{Subtract [9] - [8]: } \;0.75a \:=\:1.53 \quad\Rightarrow\quad \boxed{a \:=\:2.04}$


$\displaystyle \text{Substitute into [1]: }\;2.04 + c \:=\:2.04 \quad\Rightarrow\quad \boxed{c \:=\:0}$


. . $\displaystyle \text{Therefore: }\;h(t) \;=\;2.04\,(1.5)^t$

 

[smileyface497]

Re: Finding the equation of an exponential function from a t

Thank you, thank you! I would have never gotten that by myself.