G. Exponents and Exponential Functions
1. Exponential Functions
Def/ An exponential function is defined by: y = ax where a  0, a  1
D = Reals R =  y y  0
What problems occur:
(a) if a < 0 (negative)? Well, let ‘a’ = -4. Then y = (-4)x might look exponential but…
if x = 1/2, we would have troubles with numbers like  4   4 ?
(b) if a = 0? Well, y = 0x would just be a horizontal line (constant function) y = 0
which is not considered an exponential function. Also, there would be a ‘hole’ in the
graph at (0,0) since 00 is not defined.
(c) if a = 1? Well, consider the function y = 1x. This would
just be the horizontal line or constant function, y = 1. Well-defined but we won’t
consider this an exponential function either.
1
2
Thm/ y = ax is an increasing function iff a > 1
Thm/ y = ax is a decreasing function iff 0 < a < 1
y = 3x (incr) vs y = (1/3)x (decr)
More notes:
(i) All exponential functions contain the point (0,1).
(ii) y = 3x and y = (1/3)x (taken together) are symmetric wrt the y-axis.
(iii) y = 0 is the horizontal asymptote.
(iv) y = 3-x is equivalent to y = (1/3)x.
Power functions also involve exponents but each power function has a constant power
(exponent) and a variable base (y = xn) while an exponential function has a constant
base and a variable exponent (y = ax).
Exponential functions employ a functional notation that is looks ‘exponential’.
As a exercise, consider the possible function notation: y  exp3 x = f(x). Then
(a) f(4) = exp3(4) = 34 or 81
(b) 2  exp3(4) = 162
(c) f(x+2) = exp3(x+2) = 3x+2 = 3x  32 = exp3x + exp32 = f(x) + f(2)
Of course, f(a+b) does not in general equal f(a) + f(b).
G. Exponents and Exponential Functions (continued)
2. Exponents
Laws of Exponents (work well if the base ‘a’ is positive!)
1) a m  a n  a m  n
ex / a 3a 2   aaa  aa   a 5
am
2) n  a mn
a
a 7 aaaaaaa
 a4
ex / 3 
aaa
a
3)  a m   a mn
ex /  a 3    a 3  a 3   a 6
2
n
Also
ex / (2)0  1 but 00 is undefined.
4) a 0  1 a  0
1 1
1
ex / 42  2 
5) a  n  n
4 16
a
3 2
m m
m
ex / (3x )  9 x 6
6) (ab)  a b
(Kind of a 'distributive property of exponentiation over multiplication'?)
*In general,  a  b   a m  b m . Exponentiation does not distribute over addition.
m
ex / 32  42   3  4 
2
*All of the above statements should be qualified since we cannot divide by zero.
Although exponential functions must have positive bases (a > 0, a  1 ),
exponential expressions may have negative bases as in: (-2)3 = -8.
Also, although the above laws hold for fractional and irrational exponents, the
expressions must be well-defined. Usually the ‘fine print’ requires that the base, a, be
positive.
Ex/
a
a
1
1
2
1
1
 a 3  3 a with the restriction that a  0. Incorrect if a  0 since a 6  6 a .
6
Integer exponents have been used since the first year of algebra.
Rational exponents can be troublesome as we’ll soon see.
Let’s go back to our work with radicals to see what kind of problems we will run into.
Ex/ a  b  ab Yes? No? Well… Let’s try 4  9  36  6 ? Wrong answer!
Remember complex (imaginary) numbers? 4  9  2i  3i  6i 2  6(1)  6
Ex/  ab   a b (See Law (6) above) Yes? No? Well… not if a = -4 and b = -9.
1
Ex/
4
1
2
2
1
2
a 2  a  a Yes? No? Well… What if a = -4?
4
1
2
 4 16  2   4   4  2i (correct and imaginary)
We’d better be careful about ‘reducing fractions’ when the bases are negative!
4
 4
2
2
1
2
G. Exponents and Exponential Functions (continued)
3. Fractional Exponents and Radicals
1
First of all, think “ a n means the nth root of a .” Now let’s consider rational exponents.
Def/ a
m
n
 n a m for a  0 and m, n 
(natural numbers or positive integers)
For a  0 (positive bases), we also have a
m
n
 n am 
 a
n
m
.
So what’s the problem? Really none so long as the base is not negative.
Here’s the final word on negative bases. Don’t bother with this on a first reading.
m
 n a m for a  0 (negative base!) and for m, n (natural numbers)
IF… m is in 'lowest terms' (ie, reduced). IF NOT, all bets are off!
n
Ex/ (Just for fun!) Let’s see what trouble we can get into!
Def/ a
n
Try #1:
 8
2
Try #2:
 8
2
Try #3:
 8
2
6

6
 8
6

6
 8
6
  8  3  3 8  2 correcto-mundo! (TI-83 agrees!)
2
 6 64  2
2


6
8

2
(nope!)
 undefined (nope!)
1
*Important Ex/ So all we have to do is reduce our fractions? Right? Wrong!
Consider 4 a 4 with 'a ' possibly negative. Now if all we have to do is to reduce
4/4 to 1, then we get
4
4
a 4  a 4 ( 4 4 is not reduced so all bets are off!) . We
4
(incorrectly) set a 4  a1 . Our (incorrect) conclusion would be:
4
 3
4
 a1  3
The actual answer is of course a or in this case just '3'.
Memorize this:
2
x 2 or
x2  x
You’ll be happy to hear the final word… The power of a negative base is
UNDEFINED for irrational exponents. Try these undefined expressions
2

on your calculator:  8 = (-8)^  = ? or  8    8  ^ 2  ?
Let’s stick with positive bases and continue!
Our calculator will take care of irrational exponents. Negative exponents are easy.
Negative rational exponents are handled just like negative integer exponents.
m
1
1
1
Def/ a n  m  n a  m  n m  n m (Take your pick!) for a > 0, etc.
a
a
a n
Restrictions on variables are not always easy to spot.
Ex/ x-2 = 1/x2 x  0
5
Ex/ x 4  ? x  0 because of the 4th root
G. Exponents and Exponential Functions
3. Fractional Exponents and Radicals (continued)
Thm/ For a  0 ,
m n
 
Since a
1
n
1
m
a  mn a
a
1
mn
.
4. Equation Solving
Solving irrational equations involves the unknown quantity (usually denoted by
‘x’) under a radical sign (in the radicand).
Ex/ (Try to follow the logic of the argument carefully.)
(A) x  x  4  3x  7 We’ll move the ‘x’ over and square both sides. Right?
Notice the ‘double implication’ or equivalence here.
 x  4  2x  7
2
(B)  x  4  4x  28x  49 Now here we have only the single arrow…
(back to ‘double arrows’)
Timeout!
 0  4 x2  29 x  45
2
(now some cool factoring) When we write: ' x  2'  'stx  4' , we’re
saying, “Solutions of the 1 equation are also
 0   4 x  9  x  5 
solutions of the 2nd equation.” In terms of truth
 '4 x  9  0'  ' x  5  0' sets, “T
1st Equation  T2nd Equation”.
9
 x  x5
4
When we write: ' x2  4'  '( x  2)( x  2)  0' ,
These last four equivalence
we’re saying, “Solutions for the 1st equation
symbols indicate that the
solve the 2nd and vice versa.” In terms of truth
truth sets for these last 5
sets, “T1st Equation = T2nd Equation”.
equations are the same:
TB = 9 4 ,5 but this is not
the truth set for our original equation. We do know that any solution for A is
also a solution for B (TA  TB). So we take all the solutions to B and plug into
A: x  x  4  3x  7 . We see that TA = 5 . Do you see how and where we
got our ‘extraneous’ (extra) solution?
So how do we avoid this problem. Squaring both sides of an equation will
most often not result in an equivalent equation. Check the ‘superset’ (TB).
With other methods we really don’t need to ‘check’ our answers if we’re sure
we’ve maintained equivalence (double arrows!). Sometimes another solution
method won’t give us this problem. Here’s an easy but instructive example.
Ex/ Solve: x2  9 Method 1 leads to a common error. Method 2 is much safer.
Method 1/ Take the square root of both sides
Method 2/ Set equal to zero
 x2  9  0
 x2  3
 x  3 (nice algebra) vs x  3 (oops!)
  x  3 x  3  0 (factor)
 x  3
 x  3
Solving irrational equations (continued)
With our graphing/solver calculators we can find decimal answers easily.
What we’ll do here is learn a few weird tricks and practice our algebra.
Ex/ (A) 2 x  1  x  3  4 (With 2 radicals things can get messy.)
(standard) Method 1/ Square both sides Method 2/ Multiply by the radical conjugate
2x  1  x  3
2x  1  x  3
First isolate one radical on one side.


2x  1  4  x  3
4
Square both sides.
2 x  1  16  8 x  3  ( x  3)


2x  1  x  3

‘difference of 2 squares’ (a-b)(a+b) = a2 – b2
(2 x  1)  ( x  3)  4 2 x  1  x  3


Isolate the remaining radical on one side. Combining like terms and rewriting…
x4
2x  1  x  3 
x  12  8 x  3
4
2
 2 x  1  x  3  4 Adding to (A)
x  24 x  144  64( x  3)
x  4 x  20
Combine like terms and set equal to zero.
2 2 x  1  4  44  

4
4
2
Cross-multiply and square both sides.
x  88x  336  0
Hmm, we still have to factor this!
64(2 x  1)  x2  40 x  400
( x  4)( x  84)  0
x2  88x  336  0
x = 4, 84 (but we still have to check them)
( x  4)( x  84)  0  x = 4, 84


Either way, x  4
Ex/ (A) 3 15  2 x  3 13  2 x  4 Whoa! (Keanu Reeves imitation!) 2 cube roots!
Method 1/ “u and v” substitution?!?
Method 2/ Binomial expansion?!?
Let u  3 15  2 x and v  3 13  2 x
(a+b)3 = a3 + 3a2b + 3ab2 + b3
Then we have: u + v = 4
The plan here is to cube both sides of (A).
3
3
and u + v = 28 (= 15+2x+13-2x) where (A) is seen as ‘a + b = 4’
Factoring a ‘sum of 2 cubes’ & dividing… 15  2 x   3a 2b  3ab2  13  2 x   64
u  v  u
 uv  v 2 
u 3  v 3 28


uv
uv
4
2
2
We get: u  uv  v  7
uv4
and (A):
Solving this system (let v = 4-u)
2
u2  u  4  u    4  u   7
2
3u 2  12u  9  0  u 2  4u  3  0
(u – 1)(u – 3) = 0
u  3 15  2 x  1  x  7
u  3 13  2 x  3  x  6  x  7,6
3(a2b  ab2 )  36
ab(a+b) = 12
and (A): a+b = 4 (So divide or replace…)
ab = 3
Now cube both sides.

3
15  2 x

3
3
13  2 x

3
 27
195 – 4x – 4x2 = 27  x2  x  42  0
(x + 7)(x – 6) = 0
x  7,6
G. Exponents and Exponential Functions
4. Equation Solving (continued)
Solving exponential equations involves the unknown quantity (usually ‘x’) in
the exponent position with various constants for bases.
Ex/ The easiest exponential equation to solve is when the bases are the same.
3x = 34 iff x = 4 (Since y = 3x is a 1-1 function. See it’s graph above.)
Ex/ Another easy exponential equation to solve involves base ‘10’.
10x = 7 (It turns out that the log function, y = log10 x, is the inverse
function of exponential function, y = 10x. So all we have to do is
‘take the log’ of both sides of the equation.)
x
 log10 10 = log10 7 (log10 10x is the same as f-1(f(x)) which equals x.)
 x = log10 7 also written log 7 ,called the ‘common log’ when base is 10.
Ex/ Another easy change of base…
10x = 100  10x = 102  x = 2
Ex/ This one looks tricky…
9 x 4 x4.5  3
2
  32 
x2 4 x4.5
 32 x 8 x9  3
 2x2 + 8x – 9 = 1  2x2 + 8x – 10 = 0  x2 + 4x – 5 = 0
 (x-1)(x+5) = 0  x = 1, -5
Ex/ Here’s a fun one.
4 x  2 x1  8  0
2 
2 x
2
 2 x  21  8  0 (I know 8 = 23, but trust me… leave it alone!)
 
22 x  2  2 x  8  0  2 x
2
 2  2 x  8  0 (Do you see where we headed?)
If we make the ‘u-substitution’: u = 2x , we’ll get a quadratic in ‘2x’.
u  4 (u  2)  (2x  4)(2x  2)  0
Now 2x cannot equal -2. So that leaves 2x = 4. Answer: 2
Knowing that function values are equal may not pinpoint the domain value or
solution to an equation if the function is not one-to-one.
Ex/ Suppose f(x) = x2. Now what if x2 = 32?
Does that mean x = 3? (Answer: No!)
We did this problem earlier. Since y = x2 is not one-to-one, there may
be more than one solution here. Here we know that x = -3, 3 is correct.