More Algebra Hints
7 Natural Logarithms and Exponentials
The 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: If a > 0, b> 0 and ln(a) = ln(b) then a = b.
Inversion Properties
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ln(exp(x)) = x for all real x
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exp(ln(x)) = x if x > 0
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 ln(ab) = ln(b) +ln(a) (proof available in calculus)
Fundamental property of exponentials: exp(x1) · exp(x2) = exp(x1+x2) This follows from the uniqueness property of logarithms and the fundamental properties of logarithms.
The fundamental property of logarithms implies
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ln( 1/a) = (-1) ln(a) as 0 = ln(1) = ln ( (1/a) a )
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ln(a m) = m ln (a) for all whole numbers and then for all integers. integers.
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
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Roots and rational powers of positive numbers
How to compute using logs and exponentials
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ln(a m) = m ln (a) implies a m = exp( ln(a m)) = exp(m ln (a)). Eg (1.7)3 = exp( 5 ln(1.7)).
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Now b = a 1/m when and only when b m = a. The latter implies ln(a) = ln(b m) = m ln (b) and hence ln(b) = (1/m) ln (a). So b = exp (ln(b)) = exp( (1/m) ln(a) ) = a 1/m Eg. 8 1/3 = exp( (1/3) ln(8)
Now if m and n have no comon divisors, and n is nonzero, let the m/n power of a m/n = (a m)1/n
Then a m/n = (a m)1/n = exp( (1/n) ln(a m)) = exp( (1/n)m ln(a m)) = exp( (m/n) ln(a))
Roots and rational powers of positive numbers
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EG: 8 1/3 = exp( (1/3) ln(8)
EG 8 2/3 = exp( (2/3) ln(8)) = ( 8 1/3)
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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).
Properties of Exponentials
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.
Even Roots of Roots Numbers
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
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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
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nÖbas 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)). See above.
Odd Roots of Real Numbers
Each real number x = sign(x) |x|. For instance
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+5 = (+1) 5 as sign (5) = +1 and |+5| = 5 = distance of +5 = 5 to origin 0
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-4 = (-1) 4 as sign (4) = -1 and |-4| = 4 = distance of -4 to origin 0
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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 b1/3 and
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= sign(b) exp( (1/3) ln(|b|) ) |
is the real solution of x 3 = b. Let
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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
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nÖb= sign(b) exp( (1/n) ln(|b|) ) be the real solution of xn = b . Let
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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 |
Definition of Natural Logarithms
What the ln(x) button computes for x > 0.
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.
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