Exponentials with Arbitrary Bases Help
Introduction to Exponentials with Arbitrary Bases
We know how to define integer powers of real numbers. For instance
But what does it mean to calculate
4 or π e ?
You can calculate values for these numbers by punching suitable buttons on your calculator, but that does not explain what the numbers mean or how the calculator was programmed to calculate them. We will use our understanding of the exponential and logarithm functions to now define these exponential expressions.
If a > 0 and b is any real number then we define
To come to grips with this rather abstract formulation, we begin to examine some properties of this new notion of exponentiation:
If a is a positive number and b is any real number then
(1) ln( a b ) = b · ln a .
ln( a b ) = ln(exp( b · ln a )).
But ln and exp are inverse, so that the last expression simplifies to b · ln a .
Let a > 0. Compare the new definition of a 4 with the more elementary definition of a 4 in terms of multiplying a by itself four times.
We ordinarily think of a 4 as meaning
a · a · a · a .
According to our new definition of a b we have
a 4 = 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.
Express exp( x ) as a power of e .
According to our definition,
e x = exp( x · ln( e )).
But we learned in the last section that ln( e ) = 1. As a result,
e x = exp( x ).
You Try It : Simplify the expression ln[ e x · x e ].
Because of this last example we will not in the future write the exponential function as exp( x ) but will use the more common notation e x. Thus
Use our new definitions to simplify the expression
A = e [5·ln 2−3·ln 4] .
We next see that our new notion of exponentiation satisfies certain familiar rules. If a, d > 0 and b, c ∈ then
(i) a b + c = a b · a c
Simplify each of the expressions
You Try It : Simplify the expression ln[ e 3 x · e − y −5 · 2 4 ].
Solve the equation
( x 3 · 5) 8 = 9
for x .
You Try It : Solve the equation 4 x · 3 2 x = 7. [ Hint : Take the logarithm of both sides. See also Example 6.22 below.]
Find practice problems and solutions for these concepts at: Transcendental Functions Practice Test.
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