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Stage 5.1-3

13.04 Logarithm laws

Lesson

Logarithm laws

In the same way that there are index laws which allow us to simplify exponential expressions, there are logarithm laws that allow us to simplify logarithmic expressions. In fact, each logarithm law is a consequence of an index law.

First, consider the definition that if y=B^x then x=\log_{B}y. It follows that,\log_{B} B^x=xSubstituting x=0 and x=1 gives the following special cases:\log_{B} 1=0 \\ \log_{B} B=1

For the following proofs, we will let p=B^m and q=B^n so that \log_{B}p=m and \log_{B}q=n. Note that for any B \neq 0 there will be some values of m, \, n, \, p, and q which makes these equations true.

\displaystyle B^m \times B^n\displaystyle =\displaystyle B^{m+n}Multiplication law
\displaystyle \log_{B} (B^m \times B^n)\displaystyle =\displaystyle \log_{B} B^{m+n}Take the logarithm of both sides
\displaystyle \log_{B} (B^m \times B^n)\displaystyle =\displaystyle m+nApply \log_{B} B^x=x
\displaystyle \log_{B} pq\displaystyle =\displaystyle \log_{B}p + \log_{B}qSubstitute p=B^m and q=B^n
\displaystyle \dfrac{B^m}{B^n}\displaystyle =\displaystyle B^{m-n}Power of a power law
\displaystyle \log_{B} \dfrac{B^m}{B^n}\displaystyle =\displaystyle \log_{B} B^{m-n}Take the logarithm of both sides
\displaystyle \log_{B} \dfrac{B^m}{B^n}\displaystyle =\displaystyle m-nApply \log_{B} B^x=x
\displaystyle \log_{B} \dfrac{p}{q}\displaystyle =\displaystyle \log_{B}p - \log_{B}qSubstitute p=B^m and q=B^n
\displaystyle \left(B^m\right)^n\displaystyle =\displaystyle B^{mn}Powere of a power law
\displaystyle \log_{B} \left(B^m\right)^n\displaystyle =\displaystyle \log_{B} B^{mn}Take the logarithm of both sides
\displaystyle \log_{B} \left(B^m\right)^n\displaystyle =\displaystyle mnApply \log_{B} B^x=x
\displaystyle \log_{B} p^{n}\displaystyle =\displaystyle n \log_{B}pSubstitute p=B^m

Examples

Example 1

Simplify \dfrac{\log_{4}49}{\log_{4}7}.

Worked Solution
Create a strategy

Use the logarithm law: \log_{B} p^{n}=n \log_{B}p

Apply the idea
\displaystyle \dfrac{\log_{4}49}{\log_{4}7}\displaystyle =\displaystyle \dfrac{\log_{4}7^{2}}{\log_{4}7}Write the log as a power of 7
\displaystyle =\displaystyle \dfrac{2\log_{4}7}{\log_{4}7}Use the law: \log_{B} p^{n}=n \log_{B}p
\displaystyle =\displaystyle 2Cancel out \log_{4}7

Example 2

Simplify \log_{}2x + \log_{}50y to an expression with a single logarithmic term.

Worked Solution
Create a strategy

Use the logarithm law: \log_{B} pq= \log_{B}p + \log_{B}q

Apply the idea
\displaystyle \log_{}2x + \log_{}50y\displaystyle =\displaystyle \log_{}\left(2x \times 50y\right)Use the law: \log_{B} pq= \log_{B}p + \log_{B}q
\displaystyle =\displaystyle \log_{}\left(100xy\right)Simplify
\displaystyle =\displaystyle \log_{}100 + \log_{}xyUse the law: \log_{B} pq= \log_{B}p + \log_{B}q
\displaystyle =\displaystyle \log_{}10^2 + \log_{}xyWrite the first log as a power of 10
\displaystyle =\displaystyle 2 + \log_{}xyUse the law: \log_{B} B^x=x
Idea summary

The definition of logarithms and the index laws give us the following results:

\displaystyle \log_{B} B^x\displaystyle =\displaystyle x
\displaystyle \log_{B} 1\displaystyle =\displaystyle 0
\displaystyle \log_{B} B\displaystyle =\displaystyle 1
\displaystyle \log_{B} pq\displaystyle =\displaystyle \log_{B}p + \log_{B}q
\displaystyle \log_{B} \dfrac{p}{q}\displaystyle =\displaystyle \log_{B}p - \log_{B}q
\displaystyle \log_{B} p^{n}\displaystyle =\displaystyle n \log_{B}p

Outcomes

MA5.3-11NA

uses the definition of a logarithm to establish and apply the laws of logarithms

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