Lesson Explainer: Laws of Logarithms Mathematics

In this explainer, we will learn how to use the laws of logarithms to simplify logarithmic expressions.

Recall that given a positive number 𝑏 (provided it is ≠1) and any other positive number 𝑥, the logarithm of 𝑥 to base 𝑏 is the number 𝑦 such that 𝑏=𝑥.

We write this as log𝑥. The laws of logarithms tell us how logarithms convert products and quotients to sums and differences and also how a quotient of logarithms can be interpreted as a logarithm.

Laws of Logarithms

For a fixed base 𝑏>0, where 𝑏≠1, and any positive numbers 𝑥,𝑥, and 𝑥,

  1. products: logloglog(𝑥𝑥)=(𝑥)+(𝑥),
  2. quotients: logloglog𝑥𝑥=(𝑥)−(𝑥),
  3. powers: loglog(𝑥)=𝑝(𝑥) for any real exponent 𝑝,
  4. change of base: logloglog(𝑥)(𝑥)=𝑥.

We will remark on (4) briefly. Notice that it can also be read as logloglog(𝑥)⋅(𝑥)=(𝑥) by setting 𝑏=𝑥. This “change of base” law comes from the law of exponents that states that 𝑎=𝑎⋅ since raising 𝑥 to the expression on the left-hand side of the identity above gives us 𝑥=𝑥=𝑥=𝑥.logloglogloglog()⋅()()()()bydefinitionoflogbydefinitionoflog

This says that the number 𝑧=(𝑥)⋅(𝑥)loglog satisfies 𝑥=𝑥, which is exactly what is required for this number to be log(𝑥).

Example 1: Using the Laws of Logarithms to Evaluate Expressions

Calculate loglog192−3.

Answer

Applying the logarithm law for quotients, we have logloglogloglog192−3=1923=(64)=2=6.

Note the strategy: Apply the logarithm laws before simplifying the numeric expressions and looking for further simplifications.

Example 2: Using the Laws of Logarithms to Evaluate Expressions

Find the value of logloglog10+16−5 without using a calculator.

Answer

Applying the logarithm law for sums and then differences in turn, we have loglogloglogloglogloglogloglog10+16−5=(10⋅16)−5=(160)−5=1605=32=2=5.

Example 3: Using the Laws of Logarithms to Evaluate Expressions

Find the value of loglogloglog32+810−5 without using a calculator.

Answer

We will reduce this to a quotient of logarithms by first applying the laws for sums and differences to simplify the numerator and denominator: logloglogloglogloglogloglogloglog32+810−5=(32⋅8)10−5=256=2562.

Finally, we will apply the law that says that this itself is a logarithm: loglogloglog2562=256=2=8.

Often, the trick is to recognize the laws “in reverse”.

Example 4: Using the Laws of Logarithms to Find Equivalent Expressions

Which of the following is equal to 534+6logloglog?

  1. log3
  2. log15
  3. log243
  4. log15
  5. log243

Answer

Of course, log refers to log. We start by simplifying the numerator, using the power law of logarithms: 53=3=243.logloglog

Applying the sum law of logarithms to the denominator gives loglogloglog4+6=(4⋅6)=24.

So, 534+6=24324=243().loglogloglogloglogbychangeofbase

The correct option is (C).

Remember that the change of base law has two different statements. Note the form we use here.

Example 5: Using the Laws of Logarithms to Simplify Expressions

Simplify loglog16⋅243.

Answer

The base change result is that logloglog𝑏⋅𝑐=𝑐.

The product we have does not quite fit, but if we remember that 16=2, we can use the power rule: logloglog16=2=42.

So, logloglogloglogwhichrevealsthechangeofbaselog16⋅243=42⋅243=4243()=43=4⋅5=20.

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