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6. Exponential & Logarithmic Functions
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United States of AmericaTennessee
Algebra 2

6.05 Properties of logarithms

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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

A2.F.LE.A.1

Know the relationship between exponential functions and logarithmic functions. *

A2.MP1

Make sense of problems and persevere in solving them.

A2.MP3

Construct viable arguments and critique the reasoning of others.

A2.MP6

Attend to precision.

A2.MP8

Look for and express regularity in repeated reasoning.

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