 # Basic Logarithms A way to Undo exponents. Many things we do in mathematics involve undoing an operation.

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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

Then one day your teacher introduced you to a new symbol ─ to undo addition

3 + = 10 Could be written 10 ─ 3 =

8 – 2 =

2 + ? = 8

8 – 2 = 2 + 6 = 8

8 – 2 = 6 2 + 6 = 8

The same could be said about division ÷

40 ÷ 5 =

5 x ? = 40

40 ÷ 5 = 5 x 8 = 40

40 ÷ 5 = 8 5 x 8 = 40

Consider √49

√49 = ?

? 2 = 49

√49 = ? 7 2 = 49

√49 = 7 7 2 = 49

Exponential Equations: 5 ? = 25

Exponential Equations: 5 2 = 25

Logarithmic Form of 5 2 = 25 is log 5 25 = 2

log 5 25 = ?

5 ? = 25

log 5 25 = ? 5 2 = 25

log 5 25 = 2 5 2 = 25

Try this one…

log 7 49 = ?

7 ? = 49

log 7 49 = ? 7 2 = 49

log 7 49 = 2 7 2 = 49

and this one…

log 3 27 = ?

3 ? = 27

log 3 27 = ? 3 3 = 27

log 3 27 = 3 3 3 = 27

Remember your exponent rules? 7 0 = ? 5 0 = ?

Remember your exponent rules? 7 0 = 1 5 0 = ?

Remember your exponent rules? 7 0 = 1 5 0 = 1

log 7 1 = ?

7 ? = 1

log 7 1 = ? 7 0 = 1

log 7 1 = 0 7 0 = 1

Keep going…

log 3 1 = ?

3 ? = 1

log 3 1 = ? 3 0 = 1

log 3 1 = 0 3 0 = 1

Remember this? 1/25 = 1/ 5 2 = 5 -2

log 5 ( 1/25 )= ?

5 ? = 1/25

log 5 ( 1/25 )= ? 5 -2 = 1/25

log 5 ( 1/25 )=-2 5 -2 = 1/25

Try this one…

log 3 ( 1/81 )= ?

3 ? = 1/81

log 3 ( 1/81 )= ? 3 -4 = 1/81

log 3 ( 1/81 )=-4 3 -4 = 1/81

Let’s learn some new words. When we write log 5 125 5 is called the base 125 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 2 The argument is ___

When we write log 2 8 The base is 2 The argument is 8

Back to practice…

log 10 1000=?

10 ? =1000

log 10 1000=? 10 3 =1000

log 10 1000=3 10 3 =1000

And another one

log 10 ( 1/100 )=?

10 ? =1/100

log 10 ( 1/100)=? 10 -2 =1/100

log 10 ( 1/100)=-2 10 -2 =1/100

log 10 is used so much that we leave off the subscript (aka base)

log 10 100 can be written log 100

log 10000 =?

10 ? =10000

log 10000 =? 10 4 =10000

log 10000 = 4 10 4 =10000

And again

log 10 = ?

10 ? =10

log 10 = ? 10 1 =10

log 10 = 1 10 1 =10

What about log 33? We know 10 1 = 10 and 10 2 = 100 since 10 < 33 < 100 we 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 so log 100 < log 530 < log 1000 and thus 2 < 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.

log 4 16 = x

log 4 16 = x 4 ? = 16

log 4 16 = x 4 2 = 16

log 4 16 = x x=2 4 2 = 16

Find x in this example.

log 8 x = 2

log 8 x = 2 8 2 = ?

log 8 x = 2 8 2 = 64

log 8 x = 2 x=64 8 2 = 64

We need some rules since we want to stay in real number world. Consider log base (argument) = number The base must be > 0 The base cannot be 1 The argument must be > 0

Why can’t the base be 1? 1 4 =1 1 10 =1 That would mean  log 1 1=4  Log 1 1=10 That would be ambiguous, so we just don’t let it happen.

Why must the argument be > 0? 5 2 =25 and 25 is positive 5 0 =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.

Find x in this example.

log x 36 = 2

log x 36 = 2 x 2 = 36

log x 36 = 2 6 2 = 36 (-6) 2 = 36

log x 36 = 2 x=6 6 2 = 36 (-6) 2 = 36 -6 would make the base < 0

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