# Log Laws

```Properties of
Logarithms
Tools for solving logarithmic
and exponential equations
Let’s review some terms.
When we write
log5 125
5 is called the base
125 is called the argument
2
Logarithmic form of 5 = 25 is
log525 = 2
For all the laws
a, M and N > 0
a≠1
r is any real
Remember ln and log
ln
is a short cut for loge
log means log10
Easy ones first :
loga1 = 0
0
since a = 1
log 31= ?
log 31= ?
loga1 = 0
log 31= 0
loga1 = 0
ln 1 = ?
ln 1 = ?
loga1 = 0
ln 1 = 0
loga1 = 0
Another easy one :
logaa = 1
1
since a = a
log 55 = ?
log 55 = ?
logaa = 1
log 55= 1
logaa = 1
ln e = ?
ln e = logee = ?
ln means
loge
ln e = logee = ?
logaa = 1
ln e = 1
logaa = 1
Just a tiny bit harder :
r
logaa = r
r
r
since a = a
ln
3x
=
e
?
ln
3x
=
e
loge
3x
e
=?
ln means
loge
ln
3x
=
e
loge
3x
e
=?
log
a r
r
a
ln
3x
=
e
loge
3x
e
= 3x
log
a r
r
a
log(105y) = ?
log(105y) = ?
log means
log10
log(105y) = log10 105y = ?
log means
log10
log(105y) = log10 105y = ?
log
a r
r
a
log(105y) = log10 105y = ?
log
a r
r
a
log(105y) = log10 105y = 5y
log
a r
r
a
log a MN   log a M   log a N 
Evidence that it works (not a proof):
log
log
125   3
125   log 5  25   log 5 5 
5
5
3 
2

1
   log M   log a N 
M
a N
a
Evidence that it works (not a proof):
log
log
log
2
 25   2
125
 5   log 5 125   log 5 5 
5
5

3

1
log(2x) = ?
log(2x) = ?
log
a
 MN   log a  M   log a  N 
log(2x) = log(2) + log(x)
log
a
 MN   log a  M   log a  N 
 2 
ln 
?
 x 3
 2 
ln 
?
 x 3
M 
log a 
  log
 N 
a
 M   log a  N 
 2 
ln 
  ln 2   ln  x  3
 x 3
M 
log a 
  log
 N 
a
 M   log a  N 
Power Rule :
r
logaM = r logaM
Think of it as repeated uses
log MM   log M   log M   2 log ( M )
of
r times
a
a
a
a
ln( x )  ?
2
ln( x )  ?
2
log
M   r
r
a
log
a
M 
   2 ln( x)
ln x
2
log
M   r
r
a
log
a
M 


ln x y  ?
2
log  MN

log  M
  log  N 


ln x y  ?
2
log  MN

log  M
  log  N 


ln x y  ln( x )  ln ( y)
2
2
log  MN

log  M
  log  N 


ln x y  ln( x )  ln ( y)
2
2
log
M   r
r
a
log
a
M 


ln x y  ln( x )  ln ( y)
2
2
 2 ln( x)  ln ( y )
log
M   r
r
a
log
a
M 
NEVER DO THIS
 log

( x + y) = log(x) + log(y)
(ERROR)
WHY is that wrong?
 Log
laws tell use that
log(x) + log(y) = log ( xy)
Not log(x + y)
log MN   log M   log N 
Consider
5
=5
You know that the
and the
are equal
So if you knew that :
logaM = logaN
you would know that
M=N
And vice versa, suppose
M=N
Then it follows that
logaM = logaN
ln (x + 7) = ln(10)
ln (x + 7) = ln(10)
x+7 = 10
ln(M) = ln (N)
ln (x + 7) = ln(10)
x+7 = 10
x=3
subtract 7
log3(x + 5) = log3(2x - 4)
log3(x + 5) = log3(2x - 4)
log(M) = log(N)
log3(x + 5) = log3(2x - 4)
x+5 = 2x - 4
log(M) = log(N)
log3(x + 5) = log3(2x - 4)
x+5 = 2x - 4
9=x
oh, this step
is easy
2x
3
=
x
5
2x
3
=
x
5
If M =
then
N
ln M = ln N
2x
3
2x
ln(3 )
=
x
5
=
x
ln(5 )
If M =
then
N
ln M = ln N
2x
3
2x
ln(3 )
=
x
5
=
x
ln(5 )
log
a
M
r
 r log
a
M
2x
3
x
5
=
ln(32x) = ln(5x )
2x ln(3 ) = x ln(5)
log
a
M
r
 r log
a
M
2x
3
x
5
=
ln(32x) = ln(5x )
2x ln(3 ) = x ln(5)
simple
algebra
2x
3
x
5
=
ln(32x) = ln(5x )
2x ln(3 ) = x ln(5)
2x(ln 3) – x ln(5) = 0
simple
algebra
2x
3
x
5
=
ln(32x) = ln(5x )
2x ln(3 ) = x ln(5)
2x(ln 3) – x ln(5) = 0
x[2ln(3) – ln(5)] = 0
factor
out x
2x
3
x
5
=
ln(32x) = ln(5x )
2x ln(3 ) = x ln(5)
2x(ln 3) – x ln(5) = 0
x[2ln(3) – ln(5)] = 0
x
0
2 ln( 3)  ln( 5)
Divide out
numerical
coefficient
2x
3
x
5
=
ln(32x) = ln(5x )
2x ln(3 ) = x ln(5)
2x(ln 3) – x ln(5) = 0
x[2ln(3) – ln(5)] = 0
x
0
2 ln( 3)  ln( 5)
=0
Simplify the
fraction
log
a
M 
ln  M

ln  a 
Change of Base Formula :
When you need to
approximate log53
log a M 
ln M 
ln a 
Change of Base Formula :
When you need to
approximate log53
log 5 3 
ln 3
ln 5
Here’s one not seen as
much as some of the
others:
a
loga M
M
Here’s an example
a
log a M
e
 M
ln 3 x
 3x
```