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6. Exponential & Logarithmic Functions
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United States of AmericaFL
Grade 912

6.05 Properties of logarithms

Lesson
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Lesson

Concept summary

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

First, using the fact that \log_b (x)=n \iff x=b^n, it follows that: \log_b\left(b^x\right)=x We can also substitute x=0 and x=1 to get two special cases: \log_b\left(1\right)=0 \\\log_b\left(b\right)=1Notice that the base (of the exponent) and the base (of the logarithm) are the same.

In addition to this, we have the following laws:

  • Product law\log_b\left(xy\right) = \log_b\left(x\right)+\log_b\left(y\right)
  • Quotient law\log_b\left(\frac{x}{y}\right) = \log_b\left(x\right)- \log_b\left(y\right)
  • Power law\log_b\left(x^p\right) = p\log_b\left(x\right)
  • Change of base law\log_b\left(x\right) = \frac{\log_a\left(x\right)}{\log_a\left(b\right)}
  • Equality law\text{If }\log_b\left(x\right)=\log_b\left(y\right), \text{ then } x=y

Worked examples

Example 1

Rewrite each of the following as a sum or difference of two or more logarithms:

a

\log_{10}\left(15x\right)

Approach

We can rewrite this expression using the product law of logarithms: \log_b\left(xy\right) = \log_b\left(x\right)+\log_b\left(y\right)

15x has two prime factors, 3 and 5, and one algebraic factor, x.

Solution

\displaystyle \log_{10}\left(15x\right)\displaystyle =\displaystyle \log_{8}\left(3 \cdot 5 \cdot x\right)Express 15x by prime factors
\displaystyle =\displaystyle \log_{10} \left(3\right) + \log_{10} \left(5\right) + \log_{10} \left(x\right)Corollary of product law

Reflection

There is more than one correct answer to this question. We could express it as a sum and difference using the fact 15=\dfrac{30}{2} for instance:\log_{10}\left(15x\right)=\log_{10} \left(30\right) - \log_{10} \left(2\right) + \log_{10} \left(x\right)

b

\ln \left(\dfrac{4}{c}\right)

Approach

We can rewrite this expression using the quotient law of logarithms: \log_b\left(\frac{x}{y}\right) = \log_b\left(x\right)- \log_b\left(y\right)

Solution

\ln \left(\dfrac{4}{c}\right)= \ln \left(4\right) - \ln \left(c\right)

Example 2

Evaluate the following expressions:

a

\log_{8}\left(16\right)-\log_{8}\left(2\right)

Approach

We can rewrite this expression using the corollary of the quotient law of logarithms: \log_b\left(x\right)- \log_b\left(y\right) = \log_b\left(\frac{x}{y}\right)

Solution

\displaystyle \log_{8}\left(16\right)-\log_{2}\left(2\right)\displaystyle =\displaystyle \log_{8}\left(\frac{16}{2}\right)Corollary of quotient law
\displaystyle =\displaystyle \log_8\left(8\right)Simplify the quotient
\displaystyle =\displaystyle 1\log_b\left(b\right)=1
b

\log_{2}\left(3+\sqrt{5}\right)+\log_{2}\left(3-\sqrt{5}\right)

Approach

We can rewrite this expression using the corollary of the product law of logarithms: \log_b\left(x\right)+ \log_b\left(y\right) =\log_b\left(xy\right)

Solution

\displaystyle \log_{2}\left(3+\sqrt{5}\right)+\log_{2}\left(3-\sqrt{5}\right)\displaystyle =\displaystyle \log_{2}\left(\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)\right)Corollary of product law
\displaystyle =\displaystyle \log_2\left(9-5\right)Distributive property
\displaystyle =\displaystyle \log_2\left(4\right)Evaluate the difference
\displaystyle =\displaystyle \log_2\left(2^2\right)
\displaystyle =\displaystyle 2\log_2\left(2\right)Power law
\displaystyle =\displaystyle 2\cdot 1\log_b\left(b\right)=1
\displaystyle =\displaystyle 2

Reflection

We could also solve this, once we arrive at \log_2\left(4\right), by observing that 4=2^2. We can then use the fact x=b^n \iff \log_b x=n to get \log_2\left(4\right)=2.

Example 3

Solve for x: \log_h\left(h^{3x}\right)=30

Approach

We can rewrite this expression using the power law of logarithms: \log_b\left(x^k\right) =k\log_b\left(x\right)

Solution

\displaystyle \log_h\left(h^{3x}\right) \displaystyle =\displaystyle 30State the equation
\displaystyle 3x\log_h\left(h\right)\displaystyle =\displaystyle 30Power law
\displaystyle 3x \cdot 1\displaystyle =\displaystyle 30\log_b\left(b\right)=1
\displaystyle x\displaystyle =\displaystyle 10Divide by 3

Outcomes

MA.912.NSO.1.3

Generate equivalent algebraic expressions involving radicals or rational exponents using the properties of exponents.

MA.912.NSO.1.6

Given a numerical logarithmic expression, evaluate and generate equivalent numerical expressions using the properties of logarithms or exponents.

MA.912.NSO.1.7

Given an algebraic logarithmic expression, generate an equivalent algebraic expression using the properties of logarithms or exponents.

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