Understanding Logarithms

Version 2

Text in blue is a hyperlink.

Andy Soper

March 6, 2014

1

This document was constructed and type-set using PCTEX(a dialect of LATEX)

2

Contents

1 Introduction 41.1 Origins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Solving Exponential Equations . . . . . . . . . . . . . . . . . . . 41.3 The Inverse of an Exponential Function . . . . . . . . . . . . . . 51.4 Presenting Data with wide Dynamic Range . . . . . . . . . . . . 6

2 Just Suppose 7

3 Calculating with Logarithms I (Multiplication) 9

4 Calculating with Logarithms II - Log books and Slide-rules 11

5 Calculating with Logarithms III (Square Roots) 13

6 The Laws of Logarithms 136.1 Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136.2 Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136.3 Exponentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146.4 Restricted Identity logb(b) = 1 (See Footnote 1) . . . . . . . . . . 146.5 Restricted Identity logb(1) = 0 (See Footnote 1) . . . . . . . . . . 146.6 Raising to a Logarithmic Power . . . . . . . . . . . . . . . . . . . 146.7 Change of Base . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156.8 Restricted Identity loga(x) = 1

logx(a)

(See Footnote 1 on Page 15) . . . . . . . . . . . . . . . . . . . . . 166.9 −loga(x) = log 1

ax . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3

1 Introduction

1.1 Origins

Though out the history of mathematics arithmetic always struggled to keep pacewith mathematical developments. The problems lay with multiplication, divi-sion, raising to powers, extracting roots and trigonometric calculations. Thisproblem was substantially solved in the 17th century by John Napier’s discoveryof logarithms. Prior to this the only useful means of multiplication was ”longmultiplication” by hand.

123× 456 = ?

0 0 1 2 30 0 4 5 60 0 7 3 80 6 1 5 04 9 2 0 05 6 0 8 8

1.2 Solving Exponential Equations

Logarithms have applications far beyond arithmetic calculations and, even thoughthe electronic calculator had rendered logarithms (logs) redundant as a meansof calculation, logs remain a hugely important aspect of mathematics. Logs pro-vide the only reasonable method for accurately solving exponential equationssuch as

3x = 7

xlog(3) = log(7)

x = log(7)log(3) = 0,845

0,477 = 1, 77

Check to one decimal place,

31,77 = 7, 0

4

1.3 The Inverse of an Exponential Function

One of the most important features of logarithms is their ability to form theinverse functions of exponential functions.

f(x) = ax

y = ax

To form the inverse, reflect f(x) in y = x which requires that we exchange x andy

x = ay

log(x) = log(ay) = ylog(a)

y = log(x)loga

and

y = f−1(x) = log(x)loga

and if the base of the logarithms is ’a’

f−1(x) = logax

Figure 1: f(x) = 5x and its Inverse using base-10 (dark-blue) and base-5 (light-blue) logs

Notice how key points (red circled values) on f(x) are reflected in y = x, espe-cially when we use base-5 logs.

5

1.4 Presenting Data with wide Dynamic Range

Logarithms also provide a way to display and make sense of data that coversa wide range of values. If we wish to cover a range of salaries starting at afew thousands of Rands per year and including people earning a few hundredthousand rand to those like Bill Gates and Oprah Winfrey who earn manybillions of Rands per year.

Figure 2: Car Ownership versus Income (Ficticious)

6

2 Just Suppose

Just suppose it is the 17th century (1600-1699). Trigonometry has been knownin Europe since at least the 14th century. It is the time of the Renaissance,the rebirth of learning in Europe. Columbus has rediscovered the Americasand other great voyages of discovery are being made. Books are being printed.Copernicus and Galileo are exploring the solar system using observation andmathematics. In the middle of the century Newton and Leibniz will createmuch of modern-day differential and Integral Calculus. Science was developingrapidly and demanding more and more calculations.

There no electronic calculators and no electroniccomputers either.

Of course there aree millions of superb computerslocated between human ears!

This is what Napier found. He looked at simple arithmetic and geometric se-quences (progressions) like

Figure 3: Simple Series and 2 raised to the Arithmetic Term

He saw that the the GP values were just 2 raised to the AP values. He calledthe AP value the ”Logarithm” of the GP value to the base 2. To multiply 4 by32 we add the logarithms of 4 and 32 (2+5=7) and find that 27 = 128

Figure 4: 4× 32 = 128

7

This idea is really neat BUT there are THREE problems

1. Calculations with integers is not the problem. We need to calculate withany real positive number.

2. We need a list of logarithms of those real positive numbers. How can wecalculate a logarithm ’from first principles’?

3. The number of positive real numbers is infinite. The list will never becomplete, even if we limit the numbers to a certain number of decimalplaces.

Taking the problems in reverse order and assuming that the base is 10, and to2 places of decimals

We want . We Calculate . Resultlog10(0, 00123) = log10(0.01) + log10(0, 123) = -2.91log10(0, 0123) = log10(0.1) + log10(0, 123) = -1.910log10(0, 123) = log10(1) + log10(0, 123) = -0.910log10(1, 23) = log10(10) + log10(0, 123) = 0.0899log10(12, 3) = log10(100, 0) + log10(0, 123) = 1.0899log10(123, 0) = log10(1000, 0) + log10(0, 123) = 2.0899

Notice that the logarithm’s whole number changes by 1 when we multiply ordivine by 10. We convert the number we need the base 10 log to Scientific No-tation 0.123× 10n ’n’ is called the ’Characteristic” and the decimal part is the’Mantissa. Our table of of logarithms to 5 decimal places runs from 0,00001 to0,99999 only.

8

Now we need to calculate the actual value of log10(0.123). This will take us be-yond high school maths. Logarithms can take any value as a base. If we choosethe irrational and trancendental number 2.718281828 as base special things hap-pen. If I tell you that if

f(x) = 2.718281828x

then f ′(x) = 2.718281828x

The function is equal to its own derivative. 2.718281828 is known as thenatural constant ’e’ (after Euler) and occurs widely in nature. Logs to the basee are written as ’ln(x)’. ln(x) can be found using the Taylor series

Figure 5: Taylor Expansion for ln(x)

Finally we can find

log10(x) = ln(x)ln(10)

We have been discussing the meaning of the log of a positive real number. Whynot a negative real number? Well actually you can can find the log of a negativenumber but the answer is a Complex (non Real)number. In our world of realnumbers

logb(a) = c

Assuming that the base is positive (I have never seen a case where it was not)’c’ exponentiates ’b’ to yield ’a’.

10−3 = 0, 001

A negative exponent just inverts the argument

3 Calculating with Logarithms I (Multiplication)

Logarithms rely on the Fundamental Law of Indices

am × an = am+n

9

Suppose that

bp = a number ’r’bq = a number ’s’

’p’ is called the logarithm (log) to the base ’b’ of ’r’

logbr = p

’q’ is called the logarithm (log) to the base ’b’ of ’s’

logbs = q

We state and define as follows the Fundamental Law of Logarithms:

’p’ is the power to which ’b’ must be raised to yield ’r’ and ...

’q’ is the power to which ’b’ must be raised to yield ’s’ , Whence ...

logb(r) = p

logb(s) = q

in other words,

Log is a function which takes one parameter, the base, b, and one argument, a.The function produces a value c and

logb(a) = c

If no base is shown, b = 10 is assumed

10

Using the Fundamental Law of indices

r × s = bp × bq = bp+q = blogb(r)+logb(s)

and now, using the Fundamental Law of Logarithms

logb(r × s) = logb(r) + logb(s)

let r=123, s=456 and let the base of logs be 10

log10(123×456) = log10(123)+log10(456) = 2, 09+2, 66 = 4, 75

104,75 = 56088, 0 =⇒

4 Calculating with Logarithms II - Log booksand Slide-rules

I found the values of log(123), log (456) and 104,75 usinga calculator. How did people manage before calculators?They uses a LOT of hard hand labour to produce Tablesof Logarithms.

To find log10(123) one first divided 123 by 100.Log10(100) = 2 then log10(123) became log10(100) + log10(1, 23)or 2 + log10(1, 23)

The Table of Logarithms only had to deal with log(0,00001)to log(9,99999) for ”5-Figure Tables” To find the answer10 had to be raised to the logarithm sum. First we takethe ’4’ off the number. 104 = 10000 Now we only needto find 1075 either using the log tables ”backwards” or

11

using tables of ”anti-logs”.

Figure 6: Table of Common Logarithms - Bing Images

Figure 7: Slide Rule - Wikipedia Commons

2× 3 = 6

The graduations are positioned so that the so that theirposition corresponds to their logarithms

12

5 Calculating with Logarithms III (Square Roots)

Find√

5929 using logarithms

log10(5929) = 3, 773

Using the fundamental Laws of Logarithms and Indices

5929 = 103,773

√5929 = (103,773)

12 = 10

3,7732

log10(√

(5929)) = 3,7732 = 1

2log10(5929) = 1, 8865

101,8865 = 77, 0 to one decimal place.

Similar reasoning explains the other Laws of Logarithms

6 The Laws of Logarithms

6.1 Multiplication

logb(x) + logb(y) = logb(x× y)

6.2 Division

logb(x)− logb(y) = logb(xy )

log10(100)− log10(y) = logb(10010 )

2− 1 = 1 = log10(10)

13

6.3 Exponentiation

Y logb(x) = logb(xY )

2log10(10) = log10(10) + log10(10) = 2

2 = log10(100) = log10(102)

6.4 Restricted Identity logb(b) = 1 (See Footnote 1)

logb(b) = 1

By the Fundamental Law of Logarithms, 1 is the powerto which ’b’ must be raised to yield ’b’

6.5 Restricted Identity logb(1) = 0 (See Footnote 1)

logb(1) = 0

By the Fundamental Law of Logarithms, 0 is the powerto which ’b’ must be raised to yield ’1’

6.6 Raising to a Logarithmic Power

blogb(c) = c

Show that this is true ...

”Take logs” of both sides ...

logb(c)× logb(b) = logb(c)

but logb(b) = 1

Therefore LHS = RHS

,21An ’Identity’ is true for all values of a symbolic variable. A Restricted Identity is true

for a (restricted) range of values of a variable. In this case b ∈ R+, The set of positive Realsexcluding zero.

14

6.7 Change of Base

logp(c) =logq(c)logq(p)

Proof: Suppose I want to find y where

y = loga(x) (1)

and I only have logs to the base ’b’ available.

Using the Fundamental Law of Logarithms

x = ay (2)

Take logs of both sides to the base ’b’

logb(x) = ylogb(a) (3)

and solve for y,

y =logb(x)

logb(a)(4)

but y = loga(x) by equation (1)

loga(x) =logb(x)

logb(a)=⇒ (5)

15

6.8 Restricted Identity loga(x) = 1logx(a)

(See Footnote 1 on Page 15)

Write loga(x) using base ’x’ (See subsection 6.8 above)

loga(x) = logx(x)logx(a) = 1

logx(a)

6.9 −loga(x) = log 1ax

−loga(x) = loga(x)−1 = loga(1x) = log 1

ax

The section in red uses the Change of Base as shown insection 6.7 to change the base from ’a’ to ’1a ’ like this

loga(1x) =

log 1a( 1

x )

log 1a(a)

Let the value of denominator log 1a(a) be ∆ then

(1a)

∆ = a

and ∆ = − 1, whence

loga(1x) =

log 1a( 1

x )

−1 = −log 1a( 1

x) = log 1a(x) =⇒

16