Find the exponential function that models the data in the table below. begin{array}{|c|c|}x&-3& -2&-1&0&1&2&3& 4 y&4/27&4/9&4/3&4 &12 & 36 & 108 & 324end{array} What is the exponential regression of the data? y = square 

Introduction Given a dataset where an exponential function $y=a{b}^x$ where a and b are constants. are to be obtained. Taking logarithm on both sides of the above equation, $\log y=\log a+x\log b$ $\Rightarrow v=A+Bx$

where $v=\log y,A=\log a,B=\log b$

Hence, the exponential function is reduced to a linear function between v and x and can be solved by least square method. By principle of least squares, the normal equations for estimating A and B is given by, $\sum v=nA+B\sum x\sum vx=A\sum x+B\sum x^2$

Calculations The following is table is formed to calculate the value of A and B

$\begin{array}{ccccc}y& x& v=\log y& x^2& vx\\ 4/27& -3& -0.8293& 9& 2.4879\\ 4/9& -2& -0.3522& 4& 0.7044\\ 4/3& -1& 0.1249& 1& -0.1249\\ 4& 0& 0.6021& 0& 0\\ 12& 1& 1.0792& 1& 1.0792\\ 36& 2& 1.5563& 4& 3.1126\\ 108& 3& 2.0334& 9& 6.1002\\ 324& 4& 2.5105& 16& 10.0420\\ Total& 4& 6.7249& 44& 23.4014\end{array}$

Hence, putting the values obtained from the table to the normal equations we get, $6.7249=8A+4B......(i)23.4014=4A+44B......(ii)$

On solving the simultaneous equations, by method of comparison, it is obtained that, from $(i)A=\frac{6.7249-4B}{8}.....(iii)from(ii)A=\frac{23.4014-44B}{4}.......(iv)$

Comparing (iii) and (iv) $\frac{6.7249-4B}{8}=\frac{23.4014-44B}{4}$ Hence solving the above equation, it is obtained that, $B=0.4771$ $\Rightarrow b=Anti\log B$ $=2.9999$ Then putting the value of B in (iii), it is obtained that, $A=0.6021$ $\Rightarrow a=AntilogA$ $=4.0004$ Therefore, The required exponential regression of the data is $y=4.0004(2.9999{)}^x$