Definition

Let a, b, r \in \mathbb{R}, a, b > 0 and b ≠ 1. The logarithm of a to base b, denoted by \log_b a=r, is defined as follows.

\log_b a=r if and only if b^r=a

The number b is called the base of the logarithm, and a is called the argument.

 

Notes

Throughout this post, if the base of the logarithm is not stated, it is assumed that the base is 10. Thus, \log a = \log_{10} a. Logarithms to base 10 are called common logarithms or Briggsian logarithms. Another type of logarithm that is often used is the logarithm to base e, where e is the Euler number. It is called the natural logarithm or the Napierian logarithm. The logarithm has a special notation, namely ln.  For every a > 0, ln a = loge a.

 

Examples

log2 16 = 4 (because 24 = 16)

log5 125 = 3 (because 53 = 125)

log5 .2 = -1 (because 5-1 = .2)

log 10000 = 4 (because 104 = 10000)

ln 1 = 0 (because e0 = 1)

 

By the definition above, the base of a logarithm cannot be 1. Why? If there was not such restriction, some indeterminacy would exist. To demonstrate this, log1 1 = 4 as 14 = 1. On the other hand, it also holds that log1 1 = 2 (since 12 = 1). Then, it follows that 4 = 2, contradicting 4 ≠ 2. Therefore, the base of a logarithm must not be equal to 1.

 

Going back to the definition of logarithms, take 23 = 8. By the definition, it follows that 3 = log2 8. By substituting log2 8 for 3 in 23 = 8, it follows that 2^{\log_2 8}= 8. This is an example of how Property 1 applies.

 

Property 1

b^{\log_b a}= a      ; a, b > 0 and b ≠ 1

 

Example 1

Compute 27^{\log_3 5}.

Answer

27^{\log_3 5} =(3^3)^{\log_3 5} = (3^{\log_3 5})^3 = 5^3 = 125

 

Example 2

Compute 2^{\log_{16} 81}.

Answer

2^{\log_{16} 81}=(16^{\frac{1}{4}})^{\log_{16} 81}=(16^{\log_{16} 81})^{\frac{1}{4}}=81^{\frac{1}{4}}=3

 

Property 2

\log_b xy=\log_b x + \log_b y    ; b, x, y > 0 and b ≠ 1

 

Example 3

Suppose that log 2 = 0.3010 and log 3 = 0.4771. Compute log 6.

Answer

log 6 = log 2⋅3 = log 2 + log 3 = 0.3010 + 0.4771 = 0.7781

 

Example 4

Let log 2 = 0.3010. Compute log 2000.

Answer

log 2000 = log 2⋅1000 = log 2 + log 1000 = 0.3010 + 3 = 3.3010

 

Property 3

\log_b \frac{x}{y} = \log_b x \: - \: \log_b y    ; b, x, y > 0 and b ≠ 1

 

Example 5

Suppose that log 2 = 0.3010 and log 3 = 0.4771. Compute log 1.5.

Answer

\log 1.5 = \log \frac{3}{2} = \log 3 \: - \: \log 2 = 0.4771 \: - \: 0.3010 = 0.1761

 

Example 6

Let log 2 = 0.3010. Compute log 5.

Answer

\log 5 = \log \frac{10}{2} = \log 10 \: - \: \log 2 = 1 \: - \: 0.3010 = 0.6990

 

Property 4

\log_b a^n = n \cdot \log_b a    ; a, b > 0 and b ≠ 1 and n \in \mathbb{R}

 

Example 7

Let log 3 = 0.4771. Compute log 243.

Answer

log 243 = log 35 = 5⋅log 3 = 5⋅0.4771 = 2.3855

 

Example 8

Compute ln e√7.

Answer

ln e√7 = (√7) ⋅ ln e = √7 ⋅ 1 = √7 [Recall that ln e = loge e = 1.]

 

Example 9

Compute \ln \frac{1}{e^5}.

Answer

\ln \frac{1}{e^5}=\ln e^{-5}=(-5) \cdot \ln e=-5 \cdot 1=-5

 

Property 5

\log_{b^n} a^m= \frac{m}{n} \cdot \log_b a    ; m, n \in \mathbb{R}, n ≠ 0, a, b > 0 and b ≠ 1

 

Example 10

Compute log16 64.

Answer

\log_{16} 64=\log_{2^4} 2^6 = \frac{6}{4} \cdot \log_2 2= \frac {6}{4} \cdot 1=1.5

 

Example 11

Let log2 3 = 1.585. Compute log8 3.

Answer

\log_8 3 = \log_{2^3} 3 = \log_{2^3} 3^1 = \frac{1}{3} \cdot \log_2 3 = \frac{1 }{3} \cdot 1.585 \approx 0.5283

 

Property 6

\log_b \:a= \frac{\log_k a}{\log_k b}    ; a, b > 0 and b ≠ 1

In this formula, k is any positive real number and k ≠ 1.

 

Example 12

Compute log16 64.

Answer

\log_{16} 64 = \frac{\log_2 64}{\log_2 16}=\frac{6}{4}=1.5 (cf. Example 10 above.)

 

Example 13

Let log2 3 = 1.585. Compute log3 2.

Answer

\log_3 2=\frac{\log_2 2}{\log_2 3}=\frac{1}{1.585} \approx 0.6309

 

Property 7

\log_b a=\frac{1}{\log_a b}     ; a, b > 0 and a, b ≠ 1.

It can be easily seen that the formula in Property 7 also addresses the problem in Example 13 .

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