# Logarithms Explained

If you are familiar with the exponential function {b^N} = M then you should know that its logarithmic equivalence is {\log _b}M = N. These two seemingly different equations are in fact the same or equivalent in every way. Look at their relationship using the definition below.

## Definition of a Logarithmic Function

Let b a positive number but b \ne 1. We say {\log _b}M = N (read

**log _{b}M = N ⇔ b^{N }= M**

*the symbol **⇔** means equivalent or the same

The purpose of the equivalent equations, as shown above, is to **provide a direct link between logarithmic form and exponential form**. Understanding this basic concept can help us solve some algebra problems that require switching from one form to another.

Let’s examine further how the variables M, N and b are rearranged when the logarithmic form is expressed as exponential form, vice versa.

**Observations of the “switch” in positions of the variables:**

- The subscript
**b**in log form becomes the**base**in exponential form. - The base
**N**in log form becomes the**exponent**or superscript of**b**in exponential form. - The variable M is
**isolated**on one side of the equation.

Here are a few quick illustrations on how this logarithmic and exponential equivalence are applied.

**Logarithmic Form**

1) log_{3}81 = 4

2) log_{2}32 = 5

3) log_{5}125 = 3

4) log_{7 }49 = 2

5) log_{8 }512 = 3

6) log_{10 }100 = 2

7) log_{10 }1,000 = 3

8) log_{10 }10,000 = 4

9) log_{64 }2 = 1/6

10) log_{81 }3 = 1/4

**Exponential Form**

Since 3^{4} = 81

Since 2^{5} = 32

Since 5^{3} = 125

Since 7^{2} = 49

Since 8^{3} = 512

Since 10^{2} = 100

Since 10^{3} = 1,000

Since 10^{4} = 10,000

Since 64^{1/6} = (2^{6})^{1/6} = 2

Since 81^{1/4} = (3^{4})^{1/4} = 3

Now we are going to discuss **“special cases”** that naturally arise from the basic definition of logarithm in terms of exponential equation. We will use the definition above in order to answer some of these questions. The best way to point out the concept is through the use of examples.

#### Examples of How to Solve Basic Logarithmic Equations

**Example 1:** Solve for y in logarithmic equation **log _{3}3 = y**.

Rewriting the logarithmic equation **log _{3}3 = y** into exponential form we get

**3 = 3**. What do you think is the value of y that can make the exponential equation true? In other words, we want to find the exponent in which 3 be raised in order to get 3. It looks like the only answer is 1 because when y = 1 we have

^{y}**3 = 3**.

^{1}In fact, we can generalize this idea into a simple rule…

The logarithm of a** nonzero and nonnegative** number wherein the base is the number itself is **ALWAYS equal to 1**.

**Example 2:** Solve for x in logarithmic equation **log _{8}1 = x**.

This logarithmic equation in exponential form is written as **1 = 8 ^{x}**. What could possibly be the value of the exponent x in order to make it a true statement? Using the Zero Property of exponent,

**b**, we know that any number (exception of zero), when raised to zero, is always equal to 1. It makes perfect sense that in 1 = 8

^{0}= 1**, the value must be**

^{x}**x = 0**because

**8**

^{0}**= 1**.

There’s also a rule that handles such case when we attempt to get the logarithm of 1 with any base.

The **logarithm of 1** with any values of b ( b is a positive number but b ≠1) is **ALWAYS** equal to **zero** (0).

**Example 3:** Solve for k in logarithmic equation **log _{4}(−1) = k**.

Transforming into exponential equation, we have **−1 = 4 ^{k}**. This is like a “trick” question, right? We need to find the exponent k to make the exponential equation a true statement.

Is there really such value? I don’t think there is one! In fact, we can’t find any number that we can raise 4 by to give −1. Therefore we say that there’s no solution, or undefined.

The logarithm of a **negative number** is **undefined**.

**Example 4:** Solve for w in logarithmic equation **log _{2}(0) = w**.

The last “special case” happens when we try to find the logarithm of zero. Converting that log equation again into exponential we obtain **0 = 2 ^{w}**. We are faced again by an odd situation because we want to raise the base 2 by some exponent so that it results to zero. Is there such value? You can try. But you should agree that no such value exists! Therefore, this is another case where our answer is undefined or no solution.

The **logarithm of zero** is **undefined**.

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