Page 174 - Calculus Demystified

Transcendental Functions
CHAPTER 6
b
According to our new definition of a we have 161
4
a = exp(4 · ln a) = exp(ln a + ln a + ln a + ln a)
= exp(ln[a · a · a · a]) = a · a · a · a.
It is reassuring to see that our new definition of exponentiation is consistent
with the familiar notion for integer exponents.
EXAMPLE 6.15
Express exp(x) asa power of e.

SOLUTION
According to our definition,
x
e = exp(x · ln(e)).
But we learned in the last section that ln(e) = 1. As a result,
x
e = exp(x).
e
x
You Try It: Simplify the expression ln[e · x ].
Because of this last example we will not in the future write the exponential
x
function as exp(x) but will use the more common notation e . Thus
exp(ln x) = x becomes e ln x = x
x
ln(exp(x)) = x becomes ln(e ) = x
a+b a b
exp(a + b) =[exp(a)]·[exp(b)] becomes e = e e
exp(a) a−b e a
exp(a − b) = becomes e =
exp(b) e b
b
b
a = exp(b · ln a) becomes a = e b·ln a .
EXAMPLE 6.16
Use our new definitions to simplify the expression
A = e [5·ln 2−3·ln 4] .
SOLUTION
We write

5 3 ln 32−ln 64 e ln(32) 32 1
A = e [ln(2 )−ln(4 )] = e = = = .
e ln(64) 64 2
We next see that our new notion of exponentiation satisfies certain familiar rules.
If a, d > 0 and b, c ∈ R then
b
(i) a b+c = a · a c