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10. ln(x) and exp(x)
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7. Trig and Complex No.
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10. ln(x) and exp(x)
13. Rename the > Sign
14. Problems: Quadratics
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16. Problems: Linear Eqns I
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More Algebra Hints

7 Natural Logarithms and Exponentials - Roots and Powers

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

  • ln(exp(x)) = x for all real x

  • 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). 

PS: Search engines also send visitors to the  Exponents, Radicals & logs section for the high school precalculus view, and to ch19  for the calculus level view.    

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 

  • ln( 1/a) = (-1) ln(a) as 
     0 = ln(1) = ln ( (1/a) a )

  • 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

logc(x) = ln(x)
ln(c)
·
Here ln(e) = 1 implies loge(x) = ln(x).

The logarithm of x > 0 to a base 10 is given by

log(x) = log10(x) = ln(x)
ln(a)

The button log(x) on a calculator computes log10(x).

The definition of logc(x) in terms of ln(x) implies

  •  logc(ab) = logc(b) + logc(a) for a> 0, b > 0 and c >0

  • logc( 1/a) = (-1) logc(a) as  0 = logc(1) = logc( (1/a) a )

  • logc(a m) = m logc (a)


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Professor Whyslopes:

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Two gaps

  • The Old Algebra Gap:  Algebra  appears with too few words of explanation in high school and college mathematics.  Online Volumes 2 and 3 offer remedies.   Chapters 8 to 12 in Volume 2  put more words into the explanation and comprehension of algebra.  Chapter 14 in Volume 2 with its explicit discussion of the direct and indirect use a formulas identifies a unifying theme for mathematics and logic - all rules and patterns will be used forward and backwards. Chapters 2 to 6 and 12 to 18 in Volume 3 may further ease or avoid the very challenging use of algebra in the high level mathematics: calculus.    Calculus requires earlier high school mathematics at full strength: (i) This logically complete but long lesson on  complex numbers shows how to simplify the senior  high school exposition of circular trig functions upto to formulas in the plane  for vectors dot and cross-products.  The lesson provides the route that would have been taken in course design if the key element of the lesson, a December 2009 invention,  had been available in the 1950s.  For further algebra skill development. See the site coverage of fraction with units, proportionality,  ratios and rates, polynomials, quadratics functions  and straight line slopes and equations.
  • The Arithmetic Gap: An exact and efficient mastery of arithmetic with decimals and fractions is best (required)  for the high level  study of mathematics alone and in science, technology and business.   Pages here on arithmetic with decimals and integers,  on  fractions and solving linear equations with fractional operations on stick diagrams may help fill the gap.  That exact and efficient command should be obtained in the last years of primary school and the first years of secondary school.   

 Skill mastery in mathematics has to be seen to believed.  To that end,  learn or teach how-to write and draw the steps in mathematical figuring or  reasoning  clearly. Do not try to save space by doing a sequence of step in one place. Instead, do or record the steps in sequence on a separate lines to make each step obvious and verifiable.   


Backward Use of  A = P(1+i)n  - Derivation of formula for n 

The natural log ln(x) could be replace log(x) above. 
The algebra would be the same.

Roots and rational powers of positive numbers
How to compute using logs and exponentials

  • ln(a m) = m ln (a) implies  a m = exp( ln(a m)) = exp(m ln (a)). Egs  (1.7)3 = exp( 3 ln(1.7)) and (1.7)5 = exp( 5 ln(1.7)). 

  • 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) ) = a1/m  
    Eg. 8 1/3   = exp( (1/3) ln(8)

Now if m and n have no common 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
How to compute using logs and exponentials

  • a m = exp( ln(a m)) = exp(m ln (a))

  • a 1/m = exp( (1/m) ln(a) )

  • a m/n = (a m)1/n = exp( (m/n) ln(a))

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).

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

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

Odd Roots of Real Numbers

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 =

   _
3Öb

 = b1/3 = sign(b) exp( (1/3) ln(|b|) )

is the real solution of  x 3 = b. Let

   _
3Ö0

 = 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 numbers

Each nonzero complex number z = |z| cis(q) for some angle q with say  0 < q < 2p = 360 degrees. Put

z a = cis(qa) exp(a ln |z|) whenever a and b are real.

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.

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