Download - Basic Logarithms

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Page 1: Basic Logarithms

Basic Logarithms

A way to Undo exponents

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Many things we do in mathematics involve

undoingan operation.

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Subtraction is the

inverse of addition

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When you were in grade school, you probably learned about subtraction this way.

2 + = 8

7 + = 10

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Then one day your teacher introduced you to a new symbol

to undo addition

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3 + = 10

Could be written

10 ─ 3 =

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8 – 2 =

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8 – 2 =

2 + ? = 8

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8 – 2 =

2 + 6 = 8

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8 – 2 = 6

2 + 6 = 8

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The same could be said about division

÷

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40 ÷ 5 =

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40 ÷ 5 =

5 x ? = 40

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40 ÷ 5 =

5 x 8 = 40

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40 ÷ 5 = 8

5 x 8 = 40

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Consider √49

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= ?49

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= ?

?2 = 4949

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= ?

72 = 4949

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= 7

72 = 49

49

49

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Exponential Equations:

5? = 25

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Exponential Equations:

52 = 25

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Logarithmic Form of 52 = 25 is

log525 = 2

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log

525 = ?

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log

525 = ?

5? = 25

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log

525 = ?

52 = 25

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log

525 = 2

52 = 25

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Try this one…

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log

749 = ?

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log

749 = ?

7? = 49

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log

749 = ?

72 = 49

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log

749 = 2

72 = 49

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and this one…

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log

327 = ?

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log

327 = ?

3? = 27

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log

327 = ?

33 = 27

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log

327 = 3

33 = 27

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Remember your exponent rules?

70 = ?

50 = ?

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Remember your exponent rules?

70 = 1

50 = 1

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log

71 = ?

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log

71 = ?

7? = 1

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log

71 = ?

70 = 1

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log

71 = 0

70 = 1

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Keep going…

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log

31 = ?

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log

31 = ?

3? = 1

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log

31 = ?

30 = 1

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log

31 = 0

30 = 1

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Remember this?

1/25 = 1/ 52 = 5-2

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log

5( )= ?25

1

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5? = 1/25log

5( )= ?25

1

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5-2 = 1/25log

5( )= ?25

1

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5-2 = 1/25log

5( )= -225

1

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Try this one…

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log

3( )= ?81

1

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3? =log

3( )= ?81

1

81

1

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3-4 =log

3( )= ?81

1

81

1

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3-4 =log

3( )= -481

1

81

1

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Let’s learn some new words.

When we write log

5 125

5 is called the base125 is called the argument

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When we write log

2 8

The base is ___The argument is ___

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When we write log

2 8

The base is 2The argument is 8

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Back to practice…

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log

101000=?

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log

101000=?

10?=1000

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log

101000=?

103=1000

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log

101000=3

103=1000

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And another one

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log

10( )=?100

1

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log

10( )=?

10?= 100

1100

1

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log

10( )=?

10-2= 1001

100

1

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log

10( )=-2

10-2= 1001

100

1

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log10

is used so much that we leave

off the subscript (aka base)

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log10

100 can be written log 100

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log

10000=?

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log

10000=?

10?=10000

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log

10000=?

104=10000

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log

10000= 4

104=10000

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And again

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log

10 = ?

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log

10 = ?

10?=10

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log

10 = ?

101=10

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log

10 = 1

101=10

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What about log 33?

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What about log 33?We know 101 = 10

and 102 = 100

since 10 < 33 < 100we know

log 10 < log 33 < log 100

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Add to log 10 < log 33 < log 100

the fact that log 10 = 1

and log 100 = 2

to get

1 < log 33 < 2

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A calculator can give you an approximation of log 33. Look for the log key to find out… (okay, get it out and try)

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log 33 is approximately 1.51851394

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Guess what log 530 is close to.

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100 < 530 < 1000 solog 100 < log 530 < log 1000and thus2 < log 530 < 3

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Your calculator will tell you that log 530 ≈ 2.72427….

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Now for some practice with variables. We’ll be solving for x.

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log416 = x

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log416 = x

4? = 16

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log416 = x

42 = 16

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log416 = x

x=2 42 = 16

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Find x in this example.

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log8x = 2

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log8x = 2

82 = ?

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log8x = 2

82 = 64

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log8x = 2

x=64 82 = 64

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Find x in this example.

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logx36 = 2

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logx36 = 2

x2 = ?

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logx36 = 2

x2 = 36

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logx36 = 2

x= 6 x2 = 36

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We need some rules since we want to stay in real number world.

Consider logbase

(argument) = number

The base must be > 0 The base cannot be 1 The argument must be > 0

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Why can’t the base be 1?

14=1 110=2 That would mean

log11=4

Log11=10

That would be ambiguous, so we just don’t let it happen.

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Why must the argument be > 0?

52=25 and 25 is positive 50=1 and 1 is positive 5-2 = 1/25 and that’s positive too Since 5 to any power gives us a positive

result, the argument has to be a positive number.