|
|
|
Odds & Ends Group I Group II Would you like to show yourself or others how to be algebra power users?
|
More Algebra Hints7 Natural Logarithms and Exponentials - Roots and PowersThe natural logarithm ln(x) is defined for x > 0. The exponential function exp(x) is defined for all real x. Uniqueness (or 1 to 1) Property: Inversion Properties
For each real number a, x = exp(a) is the unique solution of a = ln(x). Solving the latter equation is one way to define or compute exp(a).
Fundamental property of logarithms Fundamental property of exponentials: The fundamental property of logarithms implies
Logarithms to base c > 0.The logarithm of x > 0 to a base c > 0 is given by
The logarithm of x > 0 to a base 10 is given by
The button log(x) on a calculator computes log10(x). The definition of logc(x) in terms of ln(x) implies
Backward Use of A = P(1+i)n - Derivation of formula for n
The natural log ln(x) could be replace log(x) above. Roots and rational powers of positive numbers
|
Roots and rational powers of positive numbers
|
EG: 8 1/3 = exp( (1/3) ln(8)
EG 8 2/3 = exp( (2/3) ln(8)) = ( 8 1/3)
2
Exponentials of Real Numbers a x = exp( x ln(a))
For x = m/n and a > 0, a x = a m/n = exp( (m/n) ln(a)) = exp( x ln(a)). This suggests putting a x = exp( x ln(a)) for x irrational. Then
a x = exp( x ln(a)) for all real x for a > 0
and not only for rational numbers. From this definition, ln a x = x ln(a). Therefore loga(a x) = x because loga(x) = ln(x)/ln(a).
Now (a x)y = exp(y ln(a x )) = exp(y x ln(a )) = a yx = a xy Therefore
(a x)y = a xy (Exponential of an exponential)
Now a xay = exp(x ln(a)) · exp(y ln(a) = exp(x ln(a)+y ln(a)) = exp( (x +y )ln(a) ) = a x+y Therefore has the exponential property
a xay = a x+y for all real numbers x and y when a > 0.
Now for the natural number e = exp(1) = 2.718281828... (irrational, deci), the natural logarithm of e, ln (e) = 1 Therefore
e x = exp( x) for all real x when a > 0
as a x = exp( x ln(a)). Calculators often have a button marked e x for the evaluation of the exponential function exp( x)
Caution: the capital EXP on some calculators will not help you with the calculation of exp(x). Use the button marked e x instead.
Here x 2 > 0 for all real numbers x. Therefore the equation x 2 = b only has solutions x when b > 0, that is only when b is non-negative. Defining
b½ =sqrt(b)
as the nonnegative real solution of x 2 = b works only if b is positive. This solution is given by a ½ = exp( ½ln(b)). See above.
Similarly, if n = 2m > 0 is an even, then x n = x 2m > 0 for all real numbers x. So the equation x 2m = b only has solutions x when b > 0, that is only when b is non-negative. The foregoing implies defining
b½m =as the 2m root of (b)
as the nonnegative real solution of x 2m = b works only if b is positive. This solution is then given by a1/n = exp( (1/n)ln(b)) where n = 2m. See above
Each real number x = sign(x) |x|. For instance
+5 = (+1) 5 as sign (5) = +1 and |+5| = 5 = distance of +5 = 5 to origin 0
-4 = (-1) 4 as sign (4) = -1 and |-4| = 4 = distance of -4 to origin 0
0 = (0)(0) as sign(0) = 0 and |0| = 0 = distance of 0 to itself.
Now sign(x) = +1, 0 or -1. In all, three cases [sign(x)]2 = 1. Therefore
x3 = [sign(x)]|x|3
The equation x3 = b = sign(b) |b| has one and only real solution real solution, namely x = sign(b) exp( (1/3) ln(|b|) ) as the horizontal line y = b intersects the graph of y = x3 at most one point. Exercise: Sketch the graph of y = x3 for -2 < x < 2. For each nonzero real number b let and
|
the cube root of b = |
_ |
= b1/3 = sign(b) exp( (1/3) ln(|b|) ) |
is the real solution of x 3 = b. Let
|
_ |
= 0 |
Similarly, if n = 2m+1 > 0 is an even, then x 2m+1 = sign(x) |x|2m+1 For each nonzero real number b let b1/n and
the n-th root of b =_
nÖb=b1/n = sign(b) exp( (1/n) ln(|b|) ) be the real solution of xn = b . Let
_
nÖ0= 0
Real roots and exponentials for complex numbersEach nonzero complex number z = |z| cis(q) for some angle q with say 0 < q < 2p = 360 degrees. Put
Then z a+ b = z a z b. Now x = z 1/n = cis(q/n) exp((1/n) ln|z|) is the so-called principal complex valued solution of the equation x n = z. (z given). But if z = b is real. Then z = |b| cis(0) or z = |b| cis( 180 degrees) with |z| = b in both cases If b > 0, then z 1/n = cis(q/n) exp((1/n) ln|z|) ) = exp((1/n) ln(b)). But if b < 0, then z 1/n = cis(180/n degrees) exp((1/n) ln|b|) ) lies on the ray with angle 180/n degrees in the first quadrant of the complex plane. This complex root does not belong to the real number line. For n odd and b < 0, it differs from the real solution x = sign(b) exp( (1/n) ln(|b|) ) of xn = b. Now x = z 1/n = cis(q/n) exp((1/n) ln|z|) is a complex valued solution of the equation x n = z = b. But the latter equation has n solutions given by the formula x = cis( (q +360 k degrees)/ n) exp((1/n) ln|z|) where 0 < k < n. When z = b < 0, the real solution or root x = sign(b) exp( (1/n) ln(|b|) ) of xn = b is not the principal complex valued solution. In the complex plane, the presence of n solutions of x n = z leads to some choice in defining or selecting the principal value of z 1/n |
Learn More: See newer treatment Algebraic,
Exponential and Logarithmic, Theory
of powers and roots in site
math forlder Exponents,
Radicals & logs
Definition of Natural Logarithms -
What the ln(x) button computes for x > 0.
A full or fuller development of natural logarithms and the exponential functions too appears in in chapter 19 of Volume 3, Why Slopes & More Math. Online, Chapter 19 has consists of three webpages 19 Logs & Powers 19 Natural Logarithms. 19 Exponential Function.
The next two diagram show the area-based definition of the natural logarithm ln(a) or ln(b) in the two mutually exclusive cases a > 1 and 0 < b < 1. Note: The graph of y=ln(x) is different from the graph of t = 1/s. The latter is used to define ln(x), not graph it.

For a ³ 1, the value of ln(a) is given by the area from s = 1 to s = a under the curve y = [1/(s)]. Here we take or assume ln(1) = 0. It can be shown that ln(a) ® 0 when when a approaches 1 through values above or greater than 1. Observe increasing a increases the area under the curve = ln(a).
For 0 < b < 1, the value of ln(b) is given by (-1) times
the area under the curve y = [1/(s)] from s = b to s
= 1.
Note: The graph of y=ln(x) is different from the graph of t = 1/s. The latter is used to define ln(x). Exercise: Find the graph of ln(x) in a text book.
Learn More:
Parents: Help
your Child/Teen Learn
Site
Math Lessons For Math
Instructors/Tutors/ Would you like to show yourself or others how to be an algebra power users? |
|
|