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Prove the second law of logarithms by using the law of exponents
Q
Answers
Answer:
Why is this true again?
Keep this in mind as you read through the proofs that follow.
Product Rule: \log_b(MN)=\log_b(M)+\log_b(N)log
b
(MN)=log
b
(M)+log
b
(N)log, start base, b, end base, left parenthesis, M, N, right parenthesis, equals, log, start base, b, end base, left parenthesis, M, right parenthesis, plus, log, start base, b, end base, left parenthesis, N, right parenthesis
Let's start by proving a specific case of the rule — the case when M=4M=4M, equals, 4, N=8N=8N, equals, 8, and b=2b=2b, equals, 2.
Substituting these values into \log_b(MN)log
b
(MN)log, start base, b, end base, left parenthesis, M, N, right parenthesis, we see:
\begin{aligned}\log_2({4\cdot 8})&=\log_2(2^2\cdot 2^3)&&\small{\gray{2^2=4\text{ and } 2^3=8}}\\ \\ &=\log_2(2^{2+3})&&\small{\gray{\text{$a^m\cdot a^n=a^{m+n}$}}}\\ \\ &=2+3&&\small{\gray{\text{$\log_b(b^c)=c$}}}\\ \\ &=\log_2(4)+\log_2(8)&&\small{\gray{\text{Since $2=\log_2(4)$ and $3=\log_2(8)$}}}\\ \end{aligned}
log
2
(4⋅8)
=log
2
(2
2
⋅2
3
)
=log
2
(2
2+3
)
=2+3
=log
2
(4)+log
2
(8)
2
2
=4 and 2
3
=8
a
m
⋅a
n
=a
m+n
log
b
(b
c
)=c
Since 2=log
2
(4) and 3=log
2
(8)
And so we have that \log_2({4\cdot 8})=\log_2(4)+\log_2(8)log
2
(4⋅8)=log
2
(4)+log
2
(8)log, start base, 2, end base, left parenthesis, 4, dot, 8, right parenthesis, equals, log, start base, 2, end base, left parenthesis, 4, right parenthesis, plus, log, start base, 2, end base, left parenthesis, 8, right parenthesis.
While this only verifies one case, we can follow this logic to prove the product rule in general.
Notice, that writing 444 and 888 as powers of 222 was key to the proof. So in general, we'd like MMM and NNN to be powers of the base bbb. To do this, we can let M=b^xM=b
x
M, equals, b, start superscript, x, end superscript and N=b^yN=b
y
N, equals, b, start superscript, y, end superscript for some real numbers xxx and yyy. Why can we do this?
Then by definition, it is also true that \log_b(M)=xlog
b
(M)=xlog, start base, b, end base, left parenthesis, M, right parenthesis, equals, x and \log_b(N)=ylog
b
(N)=ylog, start base, b, end base, left parenthesis, N, right parenthesis, equals, y.
Now we have:
\begin{aligned}\log_b(MN)&=\log_b(b^x\cdot b^y)&&\small{\gray{\text{Substitution}}}\\ \\ &=\log_b(b^{x+y})&&\small{\gray{\text{Properties of exponents}}}\\ \\ &=x+y&&\small{\gray{\text{$\log_b(b^c)=c$}}} \\\\ &=\log_b(M)+\log_b(N)&&\small{\gray{\text{Substitution}}} \end{aligned}
log
b
(MN)
=log
b
(b
x
⋅b
y
)
=log
b
(b
x+y
)
=x+y
=log
b
(M)+log
b
(N)
Substitution
Properties of exponents
log
b
(b
c
)=c
Substitution
Quotient Rule: \log_b\left(\dfrac{M}{N}\right)=\log_b(M)-\log_b(N)log
b
(
N
M
)=log
b
(M)−log
b
(N)log, start base, b, end base, left parenthesis, start fraction, M, divided by, N, end fraction, right parenthesis, equals, log, start base, b, end base, left parenthesis, M, right parenthesis, minus, log, start base, b, end base, left parenthesis, N, right parenthesis
The proof of this property follows a method similar to the one used above.
Again, if we let M=b^xM=b
x
M, equals, b, start superscript, x, end superscript and N=b^yN=b
y
N, equals, b, start superscript, y, end superscript, then it follows that \log_b(M)=xlog
b
(M)=xlog, start base, b, end base, left parenthesis, M, right parenthesis, equals, x and \log_b(N)=ylog
b
(N)=ylog, start base, b, end base, left parenthesis, N, right parenthesis, equals, y.
We can now prove the quotient rule as follows:
\begin{aligned}\log_b\left(\dfrac{M}{N}\right)&=\log_b\left(\dfrac{b^x}{ b^y}\right)&&\small{\gray{\text{Substitution}}}\\ \\ &=\log_b(b^{x-y})&&\small{\gray{\text{Properties of exponents}}}\\ \\ &=x-y&&\small{\gray{\text{$\log_b(b^c)=c$}}}\\ \\ &=\log_b(M)-\log_b(N)&&\small{\gray{\text{Substitution}}} \end{aligned}
log
b
(
N
M
)
=log
b
(
b
y
b
x
)
=log
b
(b
x−y
)
=x−y
=log
b
(M)−log
b
(N)
Substitution
Properties of exponents
log
b
(b
c
)=c
Substitution
Power Rule: \log_b(M^p)=p\log_b(M)log
b
(M
p
)=plog
b
(M)log, start base, b, end base, left parenthesis, M, start superscript, p, end superscript, right parenthesis, equals, p, log, start base, b, end base, left parenthesis, M, right parenthesis
This time, only MMM is involved in the property and so it is sufficient to let M=b^xM=b
x
M, equals, b, start superscript, x, end superscript, which gives us that \log_b(M)=xlog
b
(M)=xlog, start base, b, end base, left parenthesis, M, right parenthesis, equals, x.
The proof of the power rule is shown below.
\begin{aligned}\log_b\left(M^p\right)&=\log_b(\left({b^x}\right)^p)&&\small{\gray{\text{Substitution}}}\\ \\ &=\log_b(b^{xp})&&\small{\gray{\text{Properties of exponents}}}\\ \\ &=xp&&\small{\gray{\text{$\log_b(b^c)=c$}}}\\ \\ &=\log_b(M)\cdot p&&\small{\gray{\text{Substitution}}}\\ \\ &=p\cdot \log_b(M)&&\small{\gray{\text{Multiplication is commutative}}} \end{aligned}
log
b
(M
p
)
=log
b
((b
x
)
p
)
=log
b
(b
xp
)
=xp
=log
b
(M)⋅p
=p⋅log
b
(M)
Substitution
Properties of exponents
log
b
(b
c
)=c
Substitution
Multiplication is commutative