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:
- Problem: If a = bx, find x.
Solution: x = logba. (This is the definition
logb(x) as the inverse
- Example: If 3 = 2x, then x =
There are many more complicated problems that rely on this technique for
their solution. However, first it is necessary to isolate the
term on one side of the equation. Then you can take logarithms of both
sides to extricate the variable from the exponent.
- Problem: Suppose f(x) = keax for some
parameters k and a. Find a and k given two values f(x0) and
Solution: We have that
keax0|| = ||f(x0) |
| keax1|| = ||f(x1)|
so by dividing we get that
eax0 - ax1 = ea(x0 -
Taking logarithms we have that
Now, solving for a, we see that
|ln||------------||=ln ea(x0 -
x1) = a(x0 - x1)|
|a = ||------------------------|
Lastly, substitute your newly-found value of a, as well as x0
and f(x0) into * above, and solve for k:
|k = ||------------------|
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
f(x) = k eax, and that f(1) = 3.414 and f(3) = 6.467. Find
and k, and thus the function f.
We know that
ea|| = ||3.414|
| k e3a|| = ||6.467|
Divide to get
||= e3a -
a = e2a.|
Thus a =(1/2) ln (6.467/3.414) = .3194 ( rounded to 4 decimal
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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