More About Exponential Notation
Outline for Exponential
Notation
We introduced exponential notation, in the material on Arithmetic
Expressions, as a way of abbreviating
certain arithmetic expressions For example, we used exponentiation
notation to abbreviate 555
as
5^{3
}.
Later, in the material on algebraic
expressions we used exponential notation on variables,
writing x^{3 } as an abbreviation
for xxx. In
those situations the exponent was always a positive integer.
We will now go on to discuss how to interpret exponential notation in the
more general situation where the exponent is allowed to be any rational
number.

The law of exponents: (x
^{n})(x
^{p})
= x ^{n+p}

Some First Consequences of the law of exponents:
(x ^{n}) ^{p }= x ^{np}.
(x ^{n})/(x ^{p}) = x ^{np}

Extending the notation

1 as an exponent,
x^{1}
= x

0 as an exponent,
x^{0
}=
1.

negative exponents, x^{n}
= 1/(x^{ n})

inverses as exponents,
x^{(1/n)} = ^{n}x

fractional (or decimal) exponents, x^{n/p}
= ( ^{p}x)^{
n}

the exponential function,exp(x,
y) = x^{ y}
The Law of Exponents:
What is the result of multiplying 5^{3
}by
5^{4}
?
Well, since 5^{3}
= 555
and
5^{4}
= 5555
we
must have
(5^{3 }5^{4})
= (555)(5555)
= 5555555
= 5^{7
}=
5^{3+4} .
This example suggests that for every real number x
and all positive integers n and p
we might have the general rule
(x ^{n})(x ^{p}) = x ^{n+p}
.
If you try other examples you will see that this
rule always holds. This rule is called the law of exponents.
Some First Consequences of the law of exponents:
We can use the law of exponents to derive some other useful rules.

x^{1} = x

For any positive integer n, it makes
sense to define x^{n} = 1/x ^{n}
. This is just the extension of the law of exponents to include
negative integers for we have x^{p}
x^{n} = x ^{p}1/x
^{n}
= x ^{p}/x ^{n} = x ^{pn } = x ^{p+(n)}
.^{ } The key step here is the realization that x
^{p}/x
^{n}
= x ^{pn} follows by cancellation.
For example

x ^{5} x ^{3 } =
x x x x x/x x x = x x
x x x /x
x x = x x = x ^{2}

x ^{3} x ^{5 } =
x x x / x x x x x =
x x x / x
x x x x = 1 / x x = x ^{2}.

x^{0 }= 1, this is just
the special case where for any positive integer
n we have x^{0} = x^{nn
}=
x^{n} / x^{n} = 1.

(x ^{n}) ^{p} = x ^{np},
because x ^{n} equals
x multiplied by itself n times, while (x
^{n}) ^{p} equals x
^{n } multiplied by itself p
times and so(x ^{n}) ^{p}
equals x multiplied by itself np
times.

The above rule suggests the following rule for extending the law of exponents
to rational numbers by taking x^{(1/n)}
= ^{n}x
. This makes sense since ^{n}x
denotes the positive number, y, such
that y^{n} = x (i.e.,
y multiplied by itself n
times gives x) , but since (1/n)n
= 1, the preceding rule gives us (
x^{(1/n)} )^{n} = x, just as desired.

Generalizing we get:
x^{n/p} = ( ^{p}x)^{
n} as the meaning of an arbitrary fractional exponent n/p.
Extending the notation to arbitrary real numbers as exponents.
Exponents and Logarithms
Recall that, for any natural number n,
that the factorial of n, written as n!,is
the number n! = n(n1).(n2)
... 21.
So,
for example, 3!
= 321
= 6, 4!
= 4321
= 24, and 5!
= 54 321
= 120.
Consider the infinitely long expression
1 + (x / 1) + (x ^{2}/ 2!) +
(x ^{3} / 3!) + ... +( x ^{n}
/n!) + ...
This expression is called the MacLaurin series for the exponential function.
If we, say, plug in the value 1 for
x we get the infinitely long arithmetic expression
1 + (1 / 1) + (1 / 2!) + (1 / 3!)
+ ... +( 1 /n!) + ...
Let e_{i} denote
the result of evaluating the first i terms
of this expression . We get a sequence of numbers
e_{1} = 1
e_{2} = 1 + 1 =
2
e_{3} = 1 + 1 +
1/2! = 1 + 1 + 1/2 = 2.5
e_{4} = 1 + 1 +
1/2! + 1/3! = 1 + 1 + 1/2 + 1/6 = 2.666666....
e_{5} = 1 + 1 +
1/2! + 1/3! + 1/4! = 1 + 1 + 1/2 + 1/6 + 1/24
= 2.70833...
e_{6} = 1 + 1 +
1/2! + 1/3! + 1/4! + 1/5! = 1 + 1 + 1/2
+ 1/6 + 1/24 + 1/120 = 2.7166...
...
e_{10} = 2.7182815255731922399....
...
e_{15} = 2.7182818284582297479...
...
e_{20} = 2.7182818284590452349...
...
e_{n+1} =
e_{n} + 1/(n1)!
If you inspect this sequence of numbers you will see that as you go
up the sequence the numbers "become more and more alike"  that is

For n>1, e_{n}always
starts with 2

For n>4, e_{n}
always starts with 2.7

For n>5, e_{n}
always starts with 2.71

etc.
This suggests that the sequence is converging on (becoming
increasingly and arbitrarily close to) to some particular real number.
This is indeed the case  the number is question is called e.
The real number e is an infinite non
repeating decimal, which mean that we can not write it down completely.
But by making use of the above series we can write it down to any desired
degree of accuracy. For example

2.718281828459045235360287 (e
approximated to twentyfive decimal places).

2.7182818284590452353602874713526624977572470937000
(
e
approximated to fifty decimal places).

2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427
(e approximated to one hundred places).

Note the difference between the four digits at the end of the fiftyplace
approximation and the corresponding digits in the 100place approximation.
Why do you think they are different?
If we evaluate the expression 1 +
(x / 1) + (x ^{2}/ 2!) + (x ^{3} / 3!) + ... +( x ^{n}
/n!) + ... with x = 2,
we will get an infinite sequence of numbers which converges to
7.389056098930650227... = ee
= e^{2} . In fact, if we evaluate
the expression with x a rational
number, a/b, then the result
will be e^{a/b}.
Finally (for now), if we evaluate the expression at any real number r
then the result is e^{r}.
We can also show that if r is any
positive real number that there exists a number, let's denote
it as log(r), such that
e^{log(r)} = r. Then, given any two real
numbers, r and s
we
can compute r^{s}
by the rule r^{s }= (e^{log(r)})^{s
}=
e ^{s log(r)} . The number log(r)
is called the natural logarithm of r