Basic Logarithms
108
Basic Logarithms A way to Undo exponents
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Basic Logarithms. A way to Undo exponents. Many things we do in mathematics involve undoing an operation. Subtraction is the inverse of addition. When you were in grade school, you probably learned about subtraction this way. 2 + = 8 7 + = 10. - PowerPoint PPT Presentation
Transcript of Basic Logarithms
Basic Logarithms
A way to Undo exponents
Many things we do in mathematics involve
undoingan operation.
Subtraction is the
inverse of addition
When you were in grade school, you probably learned about subtraction this way.
2 + = 8
7 + = 10
Then one day your teacher introduced you to a new symbol
─
to undo addition
3 + = 10
Could be written
10 ─ 3 =
8 – 2 =
8 – 2 =
2 + ? = 8
8 – 2 =
2 + 6 = 8
8 – 2 = 6
2 + 6 = 8
The same could be said about division
÷
40 ÷ 5 =
40 ÷ 5 =
5 x ? = 40
40 ÷ 5 =
5 x 8 = 40
40 ÷ 5 = 8
5 x 8 = 40
Consider √49
= ?49
= ?
?2 = 4949
= ?
72 = 4949
= 7
72 = 49
49
49
Exponential Equations:
5? = 25
Exponential Equations:
52 = 25
Logarithmic Form of 52 = 25 is
log525 = 2
log
525 = ?
log
525 = ?
5? = 25
log
525 = ?
52 = 25
log
525 = 2
52 = 25
Try this one…
log
749 = ?
log
749 = ?
7? = 49
log
749 = ?
72 = 49
log
749 = 2
72 = 49
and this one…
log
327 = ?
log
327 = ?
3? = 27
log
327 = ?
33 = 27
log
327 = 3
33 = 27
Remember your exponent rules?
70 = ?
50 = ?
Remember your exponent rules?
70 = 1
50 = 1
log
71 = ?
log
71 = ?
7? = 1
log
71 = ?
70 = 1
log
71 = 0
70 = 1
Keep going…
log
31 = ?
log
31 = ?
3? = 1
log
31 = ?
30 = 1
log
31 = 0
30 = 1
Remember this?
1/25 = 1/ 52 = 5-2
log
5( )= ?25
1
5? = 1/25log
5( )= ?25
1
5-2 = 1/25log
5( )= ?25
1
5-2 = 1/25log
5( )= -225
1
Try this one…
log
3( )= ?81
1
3? =log
3( )= ?81
1
81
1
3-4 =log
3( )= ?81
1
81
1
3-4 =log
3( )= -481
1
81
1
Let’s learn some new words.
When we write log
5 125
5 is called the base125 is called the argument
When we write log
2 8
The base is ___The argument is ___
When we write log
2 8
The base is 2The argument is 8
Back to practice…
log
101000=?
log
101000=?
10?=1000
log
101000=?
103=1000
log
101000=3
103=1000
And another one
log
10( )=?100
1
log
10( )=?
10?= 100
1100
1
log
10( )=?
10-2= 1001
100
1
log
10( )=-2
10-2= 1001
100
1
log10
is used so much that we leave
off the subscript (aka base)
log10
100 can be written log 100
log
10000=?
log
10000=?
10?=10000
log
10000=?
104=10000
log
10000= 4
104=10000
And again
log
10 = ?
log
10 = ?
10?=10
log
10 = ?
101=10
log
10 = 1
101=10
What about log 33?
What about log 33?We know 101 = 10
and 102 = 100
since 10 < 33 < 100we know
log 10 < log 33 < log 100
Add to log 10 < log 33 < log 100
the fact that log 10 = 1
and log 100 = 2
to get
1 < log 33 < 2
A calculator can give you an approximation of log 33. Look for the log key to find out… (okay, get it out and try)
log 33 is approximately 1.51851394
Guess what log 530 is close to.
100 < 530 < 1000 solog 100 < log 530 < log 1000and thus2 < log 530 < 3
Your calculator will tell you that log 530 ≈ 2.72427….
Now for some practice with variables. We’ll be solving for x.
log416 = x
log416 = x
4? = 16
log416 = x
42 = 16
log416 = x
x=2 42 = 16
Find x in this example.
log8x = 2
log8x = 2
82 = ?
log8x = 2
82 = 64
log8x = 2
x=64 82 = 64
Find x in this example.
logx36 = 2
logx36 = 2
x2 = ?
logx36 = 2
x2 = 36
logx36 = 2
x= 6 x2 = 36
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
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.
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.
