# Quick Review of Exponential Equations

Remember that exponential equations all involve the independent variable in the exponent. The common types of exponential equations you will need to solve are:

1. Problem: If a = bx, find x.

Solution: x = logba. (This is the definition of logb(x) as the inverse function of bx.)

• Example: If 3 = 2x, then x = log23. There are many more complicated problems that rely on this technique for their solution. However, first it is necessary to isolate the exponential term on one side of the equation. Then you can take logarithms of both sides to extricate the variable from the exponent.

2. Problem: Suppose f(x) = keax for some parameters k and a. Find a and k given two values f(x0) and f(x1).

Solution: We have that

 keax0 = f(x0) keax1 = f(x1)

so by dividing we get that

 f(x0) eax0 ------------ = ------------ = eax0 - ax1 = ea(x0 - x1). f(x1) eax1

Taking logarithms we have that

 f(x0) ln ------------ =ln ea(x0 - x1) = a(x0 - x1) f(x1)
Now, solving for a, we see that

 ln f(x0)-ln f(x1) a = ------------------------ x0 - x1

Lastly, substitute your newly-found value of a, as well as x0 and f(x0) into * above, and solve for k:

 ln f(x0) k = ------------------ eax0

If all this looks rather complicated, remember that x0, f(x0), x1 and f(x1) are simply numbers which you will be given in the exercise.

As usual, an example with actual numbers will help.

• Example: Suppose that f is an exponential function of the form f(x) = k eax, and that f(1) = 3.414 and f(3) = 6.467. Find a and k, and thus the function f.

• Solution:

We know that

 k ea = 3.414 k e3a = 6.467

Divide to get

 6.467 e3a ------------ = -------- = e3a - a = e2a. 3.414 ea

Thus a =(1/2) ln (6.467/3.414) = .3194 ( rounded to 4 decimal places).

To find k, you can use either k = 3.414/e.3914*1 or 6.467/e.3914*3. (Try both!)

You'll find that k = 2.48 (rounded to two decimal places).

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