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=
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 = ?
49 = ?
2
? = 49
49 = ?
2
7 = 49
49
49 = 7
2
7 = 49
Exponential Equations:
?
5 = 25
Exponential Equations:
2
5 = 25
2
Logarithmic Form of 5 = 25 is
log525 = 2
log525 = ?
log525 = ?
?
5 = 25
log525 = ?
2
5 = 25
log525 = 2
2
5 = 25
Try this one…
log749 = ?
log749 = ?
?
7 = 49
log749 = ?
2
7 = 49
log749 = 2
2
7 = 49
and this one…
log327 = ?
log327 = ?
?
3 = 27
log327 = ?
3
3 = 27
log327 = 3
3
3 = 27
Remember your exponent rules?
0
7 =?
0
5 =?
Remember your exponent rules?
0
7 =1
0
5 =1
log71 = ?
log71 = ?
?
7 =1
log71 = ?
0
7 =1
log71 = 0
0
7 =1
Keep going…
log31 = ?
log31 = ?
?
3 =1
log31 = ?
0
3 =1
log31 = 0
0
3 =1
Remember this?
2
1/25 = 1/ 5 = 5
-2
1
5 25
log (
)= ?
1
5 25
log (
)= ?
?
5 = 1/25
1
5 25
log (
)= ?
-2
5 = 1/25
1
5 25
log (
)= -2
-2
5 = 1/25
Try this one…
1
log3( 81 )=
?
1
log3( 81 )=
?
1
3 =
81
?
1
log3( 81 )=
?
3
-4
1
=
81
1
log3( 81 )=
-4
3
-4
1
=
81
Let’s learn some new words.
When we write
log5 125
5 is called the base
125 is called the argument
When we write
log2 8
The base is ___
The argument is ___
When we write
log2 8
The base is 2
The argument is 8
Back to practice…
log101000=?
log101000=?
?
10 =1000
log101000=?
3
10 =1000
log101000=3
3
10 =1000
And another one
1
log10( 100 )=?
1
log10( 100 )=?
?
10 =
1
100
1
log10( 100
)=?
1
-2
10 = 100
1
log10( 100
)=-2
1
-2
10 = 100
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=?
4
10 =10000
log 10000= 4
4
10 =10000
And again
log 10 = ?
log 10 = ?
?
10 =10
log 10 = ?
1
10 =10
log 10 = 1
1
10 =10
What about log 33?
What about log 33?
1
We know 10 = 10
2
and 10 = 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.
log416 = x
log416 = x
?
4 = 16
log416 = x
2
4 = 16
log416 = x
x=2
2
4 = 16
Find x in this example.
log8x = 2
log8x = 2
2
8 =?
log8x = 2
2
8 = 64
log8x = 2
x=64
2
8 = 64
Find x in this example.
logx36 = 2
logx36 = 2
2
x =?
logx36 = 2
2
x = 36
logx36 = 2
x= 6
2
x = 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
10
1 =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
0
 5 =1 and 1 is positive
-2
 5 = 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.
