Introduction Given a dataset where an exponential function \(y=ab^x\) 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

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

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, \(A = 0.6021\) \(\Rightarrow a = Antilog A\) \(= 4.0004\) Therefore, The required exponential regression of the data is \(y = 4.0004(2.9999)^x\)