Consider the data points (4,7) and (6,2). We wish to find the exponential function that passes through these points. We could of course follow methods explored previously, transforming the y coordinates, replacing each y value by log(y) then finding the linear function log(y) = mt + b corresponding to these points and finally using the 10^x function to transform the log(y) = mt + b function into y = 10^(mt + b) = 10^b * (10^m)^t. However with two points if is often easier to simply substitute the data into the form of an exponential and solve for the parameters.

We could postulate the form y = A b^t and substitute, as we did before.  We could solve the resulting system of equations using simple algebra.

However we might wish to find the y = A (2^(kt) ) form or the y = A e^(kt) form of the exponential function. As we saw before, we could change the A b^t solution into either of the others, using 2^k = b or e^k = b. Alternatively we can substitute directly into the desired form.

Substituting into the form y = A (2^(kt)).  If we substitute the data points into the form y = A (2^(kt)) we obtain the equations

7 = A * 2^(4k)

2 = A * 2^(7k).

We will solve this equation in a manner similar to that used with the A b^t form, first dividing one equation by the other to eliminate A, solving the resulting equation for k, after which we substitute k into one of the original equations and solve for A.

Dividing the first equation by the second we obtain

3.5 = 2^(4k) / 2^(7k) = 2^(4k - 7k) = 2^(-3k).

To solve the equation 2^(-3k) = 3.5 we take the logs of both sides:

log(2^(-3k)) = log(3.5), which gives

-3k log(2) = log(3.5), so

k = log(3.5) / (-3 log(2)). Evaluating by calculator or otherwise:

k = -.602.

The first equation then reads

7 = A * 2^(4 * -.602) = A*2^(-2.408) = A * 5.31. We easily solve to obtain

A = 7 / 5.31 = 1.32.

Our model is therefore

y = 1.32 (2^(-.602 t)).

Substituting into the form y = A (e^(kt)). If we substitute the data points into the form y = A (e^(kt)) we obtain the equations

7 = A * e^(4k)

2 = A * e^(7k).

Dividing the first equation by the second we obtain

3.5 = e^(4k) / e^(7k) = e^(4k - 7k) = e^(-3k).

To solve the equation e^(-3k) = 3.5 we take natural logs of both sides:

ln (e^(-3k)) = ln (3.5), which gives

-3k ln(e) = ln(3.5), so

k = ln(3.5) / (-3 ln(e)). Evaluating by calculator or otherwise:

k = -.418.

The first equation then reads

7 = A * e^(4 * -.418) = A*e^(-1.682) = A * 5.31. We easily solve to obtain

A = 7 / 5.31 = 1.32.

Note that this is the same A value obtained in the previous example.

Our model is therefore

y = 1.32 (e^(-.418 t)).

1. Find the exponential function of the given form corresponding to each of the following pairs of points:

form A (2^(k1 t) ) thru points (-4,3) and (7,2)

form A e^(k2 t) thru points (-4,3) and (7,2)

form A b^t thru points (-4,3) and (7,2).

Check to verify that b = e^k2 = 2^k1.

2. Find the exponential function of form y = f(x) = A b^t corresponding to the points (5,3) and (10,2). Determine the values of k1 and k2 such that b = e^k2 = 2^k1 and write the corresponding functions g(x) = A (2^(k1 t)) and h(x) = A e^(k2t). So that g(7) = h(7) = f(7).

We have seen how the logarithmic model dB = 10 log(I / Io) is useful for modeling our sense of hearing, bringing and intensity range from 1 to 1,000,000,000,000 into a more comfortable scale from 0 to 120 dB. We can also model things like earthquake intensity and pH using logarithmic models. The following exercises will lead you through the essential aspects of these models:

3. The Richter scale of earthquake intensities is defined by

R =  log(I / Io),

which is identical to the dB scale except for the factor 10 in the dB scale. The difference of course is that the Richter scale measures the intensity of an earthquake rather than sound intensity.

Suppose that one earthquake measures R1 = 7.4 on the Richter scale while another measures R2 = 8.2. Then what is the ratio I2 / I1 of intensity? How many times more destructive is the 8.2 earthquake than the 7.4 earthquake? (Hint: first solve R = log(I / Io) for I. Then I1 and I2 will be the expressions you get when you substitute the corresponding values for R1 and R2.)

Find the intensity ratios for each of the following pairs of R values:

R1 = 3.2 and R2 = 6.4

R1 = 7.3 and R2 = 9.1

4. Show that if R2 = R1 + 1.6, the numerical ratio I2 / I1 can be found without known what R1 is.

If one earthquake as an R value 1.6 higher than another, what is the ratio I2 / I1?

If one earthquake as an R value 2.3 higher than another, what is the ratio I2 / I1?

If one earthquake as an R value dR higher than another, what is the ratio I2 / I1?

5. The pH of a chemical is given by

pH = - log(C)

where C is something called the hydrogen ion concentration of the substance.

Solve for C if the pH of a substance is 4.5.

Solve for C if the pH of a substance is 11.5.