(x = 2.00 , y = 0.94) and (x = 8.00, y = −1.05).Substituting the first point into the equation of the straight line, y = m x + b, gives the equation
0.94 = 2 m + b.Substituting the second point into the equation of the straight line, y = m x + b, gives the equation
−1.05 = 8 m + b.This pair of equations,
0.94 = 2 m + bis a system of two equations for the two unknowns m and b. We can solve them by elimination. Subtracting the first equation from the second equation eliminates b and gives:
−1.05 = 8 m + b
−1.99 = 6 m m = −0.33Backsubstituting this value of m into, say the first equation, 0.94 = 2 m + b, then gives 0.94 = (2)(−0.33) + b which gives b = 1.6. So the equation of the straight line is
y = −0.33 x + 1.6.Now replace back y by ln(i) and x by t:
ln (i) = −0.33 t + 1.6.Antilogging this equation gives the equation
i = e^{ −0.33 t + 1.6},or simplifying,
i = 5 e^{ −0.33 t}.
The key lesson learned from this example is that the graph of any exponential function y = a e^{ b x} is a straight line when we plot ln (y) versus x. 
The key lesson learned from this example is that the graph of any power function y = a x^{ b} is a straight line when we plot ln (y) versus ln (x). 
the semilog graph, which has a logarithmic vertical scale and a linear horizontal scale, as shown below. the loglog graph, which has a logarithmic vertical scale and a logarithmic horizontal scale, as shown below.

Q = a T ^{b}.We must find the values of a and b. To do this we will use a variation of the method described in section 6.2.
Q = 2.04 t^{ 0.558}.
P = a e^{ b t}.We must find the values of a and b. To do this we will use a variation of the method described in section 6.2.
P = 79.6 e^{ −0.461 t}.
Algebra Coach Exercises 