Explainer: Evaluating Logarithms

In this explainer, we will learn how to evaluate logarithms of different bases using laws of logarithms.

You should already know the rules of exponents and how to use them.

Logarithmic functions are the inverse of exponential functions. If the exponential function 𝑦=2 has an input of π‘₯=3, then the output will be 𝑦=9.

Now we consider the inverse of this function, which is called a logarithm: 𝑦=(π‘₯).log

Since this is the inverse of the exponential function, we can input π‘₯=9 and the output will be 𝑦=3.

This means that evaluating log(π‘š) is finding the number 𝑛 such that 𝑏=π‘šοŠ. Alternatively, we could say log(π‘š) is such that 𝑏=π‘šlog().

Next, let us look at the definition of a logarithm.

Definition: Logarithm

The logarithmic function is the inverse to the exponential function.

log(π‘š)=𝑛 means β€œπ‘ raised to the power of 𝑛 equals π‘š,” where

  • 𝑏 is the base of the logarithm,
  • π‘š is the argument,
  • 𝑛 is the exponent.

It is important to know how to change an equation from logarithmic form to exponential form, as this will aid us when solving problems involving logarithms. The essential fact is that log(π‘š)=𝑛 is equivalent to 𝑏=π‘š.

There are two special cases of logarithms where the base of the logarithm may not be obvious. These are as follows:

  • When a logarithm is written without a base, this means that the logarithm is of base 10; that is, loglog=.
  • When we have a natural logarithm, which is written as ln, this means that the base of the logarithm is 𝑒: lnlog=.

Now, we are ready to look at some examples.

Example 1: Calculating a Logarithm

What is the value of log(8)?

Answer

What this question means is β€œwhat power do we need to raise 2 to in order to obtain 8?” We can rearrange log(8)=π‘₯ into exponential form to give us 2=8, which we can solve for π‘₯.

We know that 8=2. From the rules of exponents, we know that if π‘Žβ‰ βˆ’1, 0, or 1 and π‘Ž=π‘ŽοŠο‰ then π‘š=𝑛. Therefore, π‘₯=3 and consequently log(8)=3.

Example 2: Finding the Value of a Logarithm

What is the value of log(2)?

Answer

To find the value of the logarithm log(2), we are trying to solve the equation log(2)=π‘₯.

We can convert this equation into exponential form to give us ο€Ό12=2, and now we need to solve for π‘₯. Using the rules of exponents, we know that π‘Ž=1π‘ŽοŠ±ο—ο—. Therefore, ο€Ό12=(2)=2.ο—οŠ±ο—

Hence, we can see that βˆ’π‘₯=1 which gives π‘₯=βˆ’1. From this we conclude that log(2)=βˆ’1.

Before we go into some more challenging examples, let us state the rules of logarithms.

The Rules of Logarithms

For any base 𝑏>0, π‘š>0, 𝑛>0, and real number π‘₯, the following rules apply:

  1. Logarithms of 1: log(1)=0.
  2. Logarithms of the base, 𝑏: log(𝑏)=1.
  3. Logarithms of powers: loglogοŒ»ο—οŒ»(π‘š)=π‘₯(π‘š).
  4. Logarithms of products: logloglog(π‘šπ‘›)=(π‘š)+(𝑛).
  5. Logarithms of quotients:logloglogοŒ»οŒ»οŒ»ο€»π‘šπ‘›ο‡=(π‘š)βˆ’(𝑛).

Given that we know how to convert between logarithms and exponentials, it can be useful to see the link between the rules of logarithms and the corresponding rules for exponents.

  1. The logarithms of 1 rule is equivalent to the zero exponent rule: 𝑏=1.
  2. The logarithm of the base rule is equivalent to the exponent rule: 𝑏=𝑏.
  3. The logarithms of powers rule does not have a corresponding exponent rule. However, we can see how it is derived algebraically. Using the definition of logarithms, we have that 𝑏=𝑛.log() Then, if we raise both sides to the power of π‘š, we obtain 𝑏=𝑛.()log But, we can also use the definition of logarithms on 𝑛, which gives us 𝑛=𝑏.()log From this, we see that 𝑏=𝑏,()()loglog which implies the logarithms of powers rule.
  4. The logarithms of products rule is equivalent to the product rule for exponents: 𝑏⋅𝑏=π‘οŠο‰οŠοŠ°ο‰.
  5. The logarithms of quotients rule is equivalent to the quotient rule for exponents: 𝑏𝑏=π‘οŠο‰οŠοŠ±ο‰.

Using these rules of logarithms, we will consider some more examples.

Example 3: Evaluating a Logarithm

What is the value of logοŠ¨ο€Ό1128?

Answer

First, let us note that we can write 1128=128. From this, we can use our logarithm of powers rule to obtain logloglogοŠ¨οŠ¨οŠ±οŠ§οŠ¨ο€Ό1128=ο€Ή128=(βˆ’1)(128).

Next, we can use the fact that 128=2 and substitute this into our logarithm; again, we can use the rule for logarithms of powers to get βˆ’(128)=βˆ’ο€Ή2=βˆ’7(2).logloglog

Finally, we can use the rule for logarithms of the base to obtain that log(2)=1. From this, we get the solution of logοŠ¨ο€Ό1128=βˆ’7.

Example 4: Using Logarithm Rules to Calculate Logarithms

Calculate 2(4)+7(13)loglog, giving your answer to the nearest thousandth.

Answer

Using the logarithms of powers rule, we get that 2(4)+7(13)=ο€Ή4+ο€Ή13.loglogloglog

Next, we use the logarithm of products rule to combine these into one logarithm, giving us logloglogο€Ή4+ο€Ή13=ο€Ή4Γ—13. Now, note that if the logarithm is written without a base, this means the base is 10: loglog=. In order to get our solution, we just need to calculate this logarithm. We get logοŠ§οŠ¦οŠ¨οŠ­ο€Ή4Γ—13=9.001723449…. Finally, we need to round to the nearest thousandth, giving us an answer of 2(4)+7(13)=9.002.loglog

Example 5: Evaluating a Logarithmic Expression

Find the value of loglogloglogοŠͺο€Ήο€Ήπ‘₯ο…ο…βˆ’ο€Ήο€Ήπ‘₯.

Answer

Using the logarithms of quotients rules, we have that logloglogloglogloglogοŠͺοŠͺο€Ήο€Ήπ‘₯ο…ο…βˆ’ο€Ήο€Ήπ‘₯=ο€Ώο€Ήπ‘₯(π‘₯). Now, we use the logarithms of powers rule to simplify this: logloglogloglogloglogοŠͺοŠ¨οŠ¨ο€Ώο€Ήπ‘₯(π‘₯)=ο€½32(π‘₯)4(π‘₯)=(8).

Next, we us the fact that 8=2 to obtain loglog(8)=ο€Ή2.

Finally, we use both the logarithms of powers and the logarithms of the base rules to get a solution of loglogloglogloglogοŠͺοŠ¨οŠ©οŠ¨ο€Ήο€Ήπ‘₯ο…ο…βˆ’ο€Ήο€Ήπ‘₯=ο€Ή2=3(2)=3.

Let us finish by highlighting some key points.

Key Points

  1. The logarithmic function is the inverse of the exponential function. Equivalently, evaluating log(π‘š) is finding the number 𝑛 such that 𝑏=π‘šοŠ.
  2. The two cases where the base of the logarithm is not obvious are loglog= and lnlog=.
  3. Logarithms of 1: log(1)=0.
  4. Logarithms of the base: log(𝑏)=1.
  5. Logarithms of a power: loglog(π‘š)=𝑛(π‘š).
  6. Logarithms of a product: logloglog(π‘šπ‘›)=(π‘š)+(𝑛).
  7. Logarithms of a quotient: logloglogοŒ»οŒ»οŒ»ο€»π‘šπ‘›ο‡=(π‘š)βˆ’(𝑛).

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