Unit 11 · Lesson 3 of 5

Properties of logarithms

About 25 minutes · 5 steps · one idea: log rules are exponent rules in disguise

You'll be able to: expand one log into several, condense several logs into one, and use change of base on a calculator.

  1. 1 Warm-up
  2. 2 The idea
  3. 3 Examples
  4. 4 Practice
  5. 5 Wrap-up

Step 1 of 5Warm-up

Two quick questions from earlier lessons. Answer from memory first, then check.

\(\log_2 \dfrac{1}{8}\)

2 to what power is \(\dfrac18\)? \(8 = 2^3\), so \(\dfrac18 = 2^{-3}\). Answer: \(-3\).

Simplify \(2x^3 \cdot 3x^4\).

Multiply the numbers: \(2 \cdot 3 = 6\). Same base, so add the exponents: \(x^{3+4} = x^7\). Answer: \(6x^7\).

Step 2 of 5The idea

A log is an exponent. When you multiply powers, you add exponents. So the log of a product is a sum of logs. The other rules come the same way.

Product rule. \(\log_b(MN) = \log_b M + \log_b N\)

Quotient rule. \(\log_b\left(\dfrac{M}{N}\right) = \log_b M - \log_b N\)

Power rule. \(\log_b\left(M^p\right) = p \cdot \log_b M\)

Change of base. \(\log_b x = \dfrac{\ln x}{\ln b}\) (or \(\dfrac{\log x}{\log b}\)). Use it to get a decimal on a calculator.

Expand means break one log into several: times becomes plus, divide becomes minus, a power moves to the front.

Condense means go backward into one log: first move each number in front up as a power, then plus becomes times and minus becomes divide.

Why this shows up on the placement test: "which expression is equivalent" log questions are common, and you need to condense logs to solve log equations.

Step 3 of 5Worked examples

Example 1: expand \(\log_3\left(\dfrac{9x^2}{y}\right)\)

  1. Quotient rule: \(\log_3(9x^2) - \log_3 y\). Divide becomes minus.
  2. Product rule: \(\log_3 9 + \log_3(x^2) - \log_3 y\). \(9x^2\) is 9 times \(x^2\).
  3. Power rule: \(\log_3 9 + 2\log_3 x - \log_3 y\).
  4. \(\log_3 9 = 2\) because \(3^2 = 9\). Final: \(2 + 2\log_3 x - \log_3 y\).

Example 2: condense \(3\ln x + \ln 2 - \ln(x+1)\), then a change of base

  1. Powers first: \(3\ln x = \ln\left(x^3\right)\). The 3 in front moves up as an exponent.
  2. Plus becomes times: \(\ln\left(x^3\right) + \ln 2 = \ln\left(2x^3\right)\).
  3. Minus becomes divide: \(\ln\left(\dfrac{2x^3}{x+1}\right)\).
  4. Change of base (calculator OK): \(\log_5 20 = \dfrac{\ln 20}{\ln 5} \approx \dfrac{2.996}{1.609} \approx 1.86\). Check: \(5^{1.86}\) is close to 20.

Step 4 of 5Practice

Type your answer and press Check. If you miss twice, open the worked solution. Then mark how it went. Use a calculator only where the problem says so.

Expand: \(\log_2\left(x^5 y\right) = \square \cdot \log_2 x + \log_2 y\). What number goes in the box?

Product rule: \(\log_2\left(x^5\right) + \log_2 y\).

Power rule: \(5\log_2 x + \log_2 y\). Answer: \(5\).

Which is the expanded form of \(\log\left(\dfrac{x^2}{y z^3}\right)\)?

  1. \(2\log x - \log y + 3\log z\)
  2. \(2\log x - \log y - 3\log z\)
  3. \(2\log x + \log y - 3\log z\)
  4. \(\log 2x - \log y - \log 3z\)

Quotient rule: \(\log\left(x^2\right) - \log\left(yz^3\right)\). The whole bottom is subtracted.

Product rule on the bottom, keeping it in parentheses: \(\log\left(x^2\right) - \left(\log y + \log\left(z^3\right)\right) = \log\left(x^2\right) - \log y - \log\left(z^3\right)\).

Power rule: \(2\log x - \log y - 3\log z\). Choice B. (Choice A forgot to distribute the minus; D put the powers in the wrong place.)

Expand and evaluate any log you can: \(\log_2(8x) = \square + \log_2 x\). What number goes in the box?

Product rule: \(\log_2 8 + \log_2 x\).

\(\log_2 8 = 3\) because \(2^3 = 8\). Answer: \(3\).

Condense \(\log x + \log y - \log z\) into one log, \(\log(M)\). What is \(M\)?

Plus becomes times: \(\log x + \log y = \log(xy)\).

Minus becomes divide: \(\log\left(\dfrac{xy}{z}\right)\). Answer: \(M = \dfrac{xy}{z}\).

Condense \(2\ln x + 3\ln y\) into one log, \(\ln(M)\). What is \(M\)?

Powers first: \(2\ln x = \ln\left(x^2\right)\) and \(3\ln y = \ln\left(y^3\right)\).

Plus becomes times: \(\ln\left(x^2y^3\right)\). Answer: \(M = x^2y^3\).

Condense \(3\log_2 x - 2\log_2(x+1)\) into one log, \(\log_2(M)\). What is \(M\)?

Powers first: \(\log_2\left(x^3\right) - \log_2\left((x+1)^2\right)\).

Minus becomes divide: \(\log_2\left(\dfrac{x^3}{(x+1)^2}\right)\). Answer: \(M = \dfrac{x^3}{(x+1)^2}\).

Find the exact value: \(\log_6 4 + \log_6 9\).

Neither log is a whole number alone, so condense first: \(\log_6(4 \cdot 9) = \log_6 36\).

\(6^2 = 36\), so the answer is \(2\).

Calculator OK. Use change of base to find \(\log_3 20\). Round to the nearest hundredth.

\(\log_3 20 = \dfrac{\ln 20}{\ln 3} \approx \dfrac{2.9957}{1.0986} \approx 2.7268\).

Rounded: \(2.73\). Check: \(3^{2.73}\) is about 20, and it makes sense that it is between 2 and 3, since \(3^2 = 9\) and \(3^3 = 27\).

Step 5 of 5Wrap-up

You can now use the product, quotient, and power rules to expand or condense logs, and turn any log into a decimal with \(\log_b x = \dfrac{\ln x}{\ln b}\).

Read or watch next: OpenStax Intermediate Algebra 2e, 10.4 Use the Properties of Logarithms (free textbook, with more practice at the end of the section).

Also good: Khan Academy: Properties of logarithms (short videos). Quick rules: Exponentials and logs cheat sheet.

Something unclear? Ask your teacher bot a follow-up question, for example: "Why does the log of a product turn into a sum?" It knows this course and can explain things another way.