
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 = b^{x}, find x.
Solution: x = log_{b}a. (This is the definition
of
log_{b}(x) as the inverse
function
of b^{x}.)
 Example: If 3 = 2^{x}, then x =
log_{2}3.
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.
 Problem: Suppose f(x) = ke^{ax} for some
parameters k and a. Find a and k given two values f(x_{0}) and
f(x_{1}).
Solution: We have that
ke^{ax0}  =  f(x_{0}) 
ke^{ax1}  =  f(x_{1}) 
so by dividing we get that
f(x_{0})   e^{ax0} 
  =    =
e^{ax0  ax1} = e^{a(x0 
x1).} 
f(x_{1})  
e^{ax1} 
Taking logarithms we have that
 f(x_{0})  
ln    =ln e^{a(x0 
x1)} = a(x_{0}  x_{1}) 
 f(x_{1})  
Now, solving for a, we see that
 ln f(x_{0})ln
f(x_{1}) 
a =   
 x_{0} 
x_{1} 
Lastly, substitute your newlyfound value of a, as well as x_{0}
and f(x_{0}) into * above, and solve for k:
 ln
f(x_{0}) 
k =   
 e^{ax0} 
If all this looks rather complicated, remember that x_{0},
f(x_{0}), x_{1} and f(x_{1}) 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 e^{ax}, 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
e^{a}  =  3.414 
k e^{3a}  =  6.467 
Divide to get
6.467   e^{3a} 
  =  
 = e^{3a 
a} = e^{2a.} 
3.414   e^{a} 
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).
(back to exponential functions)
(to Field Guide hub)
