If you understand what an exponent is, you can very quickly see why the three rules of exponents work. But why do logarithms have these three properties?
As you work through the text, you will demonstrate these rules intuitively, by viewing the logarithm as a counter. (
log28log28 size 12{"log" rSub { size 8{2} } 8} {} asks “how many 2s do I need to multiply, in order to get 8?”) However, these rules can also be rigorously proven, using the laws of exponents as our starting place.
Table 2
Proving the First Law of Logarithms,
logx(ab)=logxa+logxblogx(ab)=logxa+logxb size 12{"log" rSub { size 8{x} } \( ital "ab" \) ="log" rSub { size 8{x} } a+"log" rSub { size 8{x} } b} {} 
m
=
log
x
a
m
=
log
x
a
size 12{m="log" rSub { size 8{x} } a} {}

I’m just inventing
m
m
size 12{m} {}
to represent this log

x
m
=
a
x
m
=
a
size 12{x rSup { size 8{m} } =a} {}

Rewriting the above expression as an exponent. (
log
x
a
log
x
a
size 12{"log" rSub { size 8{x} } a} {}
asks “
x
x
size 12{x} {}
to what power is
a
a
size 12{a} {}
?” And the equation answers: “
x
x
size 12{x} {}
to the
m
m
size 12{m} {}
is
a
a
size 12{a} {}
.”)

n
=
log
x
b
n
=
log
x
b
size 12{n="log" rSub { size 8{x} } b} {}

Similarly,
n
n
size 12{n} {}
will represent the other log.

x
n
=
b
x
n
=
b
size 12{x rSup { size 8{n} } =b} {}


log
x
(
ab
)
=
log
x
(
x
m
x
n
)
log
x
(
ab
)
=
log
x
(
x
m
x
n
)
size 12{"log" rSub { size 8{x} } \( ital "ab" \) ="log" rSub { size 8{x} } \( x rSup { size 8{m} } x rSup { size 8{n} } \) } {}

Replacing
a
a
size 12{a} {}
and
b
b
size 12{b} {}
based on the previous equations

=
log
x
(
x
m
+
n
)
=
log
x
(
x
m
+
n
)
size 12{ {}="log" rSub { size 8{x} } \( x rSup { size 8{m+n} } \) } {}

This is the key step! It uses the first law of exponents. Thus you can see that the properties of logarithms come directly from the laws of exponents.

=
m
+
n
=
m
+
n
size 12{ {}=m+n} {}

=
log
x
(
x
m
+
n
)
=
log
x
(
x
m
+
n
)
size 12{ {}="log" rSub { size 8{x} } \( x rSup { size 8{m+n} } \) } {}
asks the question: “
x
x
size 12{x} {}
to what power is
x
m
+
n
x
m
+
n
size 12{x rSup { size 8{m+n} } } {}
?” Looked at this way, the answer is obviously
(
m
+
n
)
(
m
+
n
)
size 12{ \( m+n \) } {}
. Hence, you can see how the logarithm and exponential functions cancel each other out, as inverse functions must.

=
log
x
a
+
log
x
b
=
log
x
a
+
log
x
b
size 12{ {}="log" rSub { size 8{x} } a+"log" rSub { size 8{x} } b} {}

Replacing
m
m
size 12{m} {}
and
n
n
size 12{n} {}
with what they were originally defined as. Hence, we have proven what we set out to prove.

To test your understanding, try proving the second law of logarithms: the proof is very similar to the first. For the third law, you need invent only one variable,
m=logxam=logxa size 12{m="log" rSub { size 8{x} } a} {}. In each case, you will rely on a different one of the three rules of exponents, showing how each exponent law corresponds to one of the logarithms laws.
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