# Logarithms: Properties, Logarithmic Equations, Conditions, Hills, Problems

**Logarithm** is a mathematical operation where this operation is the operation of the inverse (or inverse) of the exponent or power. The base or principal in this logarithmic formula is generally in the form of the letter a.

Or there is also a mention if this logarithm is an inverse or the inverse of the power (exponent) used in **determine the exponent of a base number**.

In English, the logarithm is called

logarithm.

So in essence, by studying logarithms, we can find the power of a number whose power is known.

Table of contents

## Logarithm

After you know what a logarithm is, then you are also obliged to know the general form of this logarithm.

Here is the general form of the logarithm:

The general form of the logarithm:

**If a ^{n} = x then ^{a}logx = n**

**Information:**

a: is the basis, which has the following conditions: a > 0 and a 1.

x: is the number that the algorithm is looking for (numerus), the conditions are: x > 1

n: is the power of the logarithm.

Now is the time for you to look at the example questions below so that you can better understand the description above:

- When 3
^{2}= 9, then in logarithmic form it will change to^{3}log 9 = 2 - When 2
^{3}= 8, then in logarithmic form it will change to^{2}log 8 = 3 - When 5
^{3}= 125, then in logarithmic form it will change to^{5}log 125 = 3

How are you? Now I'm starting to understand *right*?

*Well*, usually *here*, you will still often experience confusion in determining which number is the base and which number is the numerus.

Logarithmis a mathematical operation where is the inverse of the exponent or power.The basic formula of the logarithm: b

^{c }= a is written as^{b}log a = c (b is called the base logarithm).

Is not it?

Calm down guys, the key that you just have to remember is if **base number** It is **base**, **located at the top before the 'log' sign. **And **number****rank result** it is called as **numerus**, **located at the bottom after the word 'log'**. Easy *right*?

## Logarithmic Equations

**Logarithmic equation****a** is an equation in which the variable is the base of the logarithm.

This logarithm can also be defined as a mathematical operation which is the inverse (or inverse) of the exponent or a power.

### Example Number

Here we will give some examples of logarithmic numbers, including the following:

Rank | Logarithmic Example |

2^{1} = 2 |
^{2}log 2 = 1 |

2^{0} = 1 |
^{2}log 1 = 0 |

2^{3} = 8 |
^{2}log 8 = 3 |

2-^{3} = 8 |
^{2}logs = -3 |

9^{3/4} = 3√3 |
^{9}log 3√3 = 3/4 |

10^{3} = 1000 |
log 1000 = 3 |

Next, logarithms also have some properties that **Required** for you to understand, *here*. Why mandatory?

This is because these characteristics will later become your provision in working on logarithmic problems with ease.

Without understanding the properties of logarithms, you will not be able to work on logarithm problems, *you know*!

Then, anything *the hell* What are the properties of the logarithm? *Come on*, note the reviews below.

## Logarithmic Properties

The following are some of the properties of logarithms that you must understand, including:

loga = 1 |

log 1 = 0 |

log aⁿ = n |

log bⁿ = n • log b |

log b • c = log b + log c |

log b/c = log b – log c |

log b m = m/n • log b |

log b = 1 b log a |

log b • b log c • c log d = log d |

log b = c log b c log a |

In addition to some of the properties above, there are also some properties of logarithmic equations, including:

## Properties of Logarithmic Equations

The logarithmic equation also has some special properties, these properties are as follows:

**1. Logarithmic Properties of Multiplication **

The logarithmic property of multiplication is a result of the addition of two other logarithms in which the value of the two numerals is a factor of the initial numeric value.

^{a}logs p. q = ^{a}log p + ^{a}log q

There are several conditions for this one trait, namely: a > 0, a \ne 1, p > 0, q > 0.

**2. Logarithmic Multiplication**

Multiplication of logarithms is a property of logarithm a which can be multiplied by logarithm b if the numeric value of logarithm a is equal to the base number of logarithm b.

The result of the multiplication is a new logarithm with the base number equal to logarithm a. And has the same numeric value as logarithm b.

^{a}log b x ^{b}logc = ^{a}log c

There are several conditions for this one trait, namely: a > 0, a \ne 1.

**3. Nature of Division **

The logarithmic property of division is the result of subtracting two other logarithms where the value of the two numerals is a fraction or division of the initial logarithm numeric value.

^{a}log p/q: ^{a}log p – ^{a}log q

There are several conditions for this one trait, namely: a > 0, a \ne 1, p > 0, q > 0.

**4. Inversely Comparable Traits**

The inversely proportional logarithm property is a property with other logarithms that have the base number value and the numerus interchangeable.

^{a}logb = 1/^{b}log a

There are several conditions for this one trait, namely: a > 0, a \ne 1.

**5. Opposite Sign **

The logarithmic property of opposite sign is a property with a logarithm whose numerus is an inverse fraction of the initial logarithm numeric value.

^{a}log p/q = – ^{a}log p/q

There are several conditions for this one trait, namely: a > 0, a \ne 1, p > 0, q > 0.

**6. Nature of Powers **

The logarithmic property of powers is a property whose numeric value is an exponent. And can be used as a new logarithm by issuing the power to a multiplier.

^{a}log b^{p} = p. ^{a}log b

There are several conditions for this one trait, namely: a > 0, a \ne 1, b > 0

**7. Power of Logarithmic Principal Numbers **

The power of a logarithmic power of a base number is a property where the value of the base number is a exponent (power) that can be used as a new logarithm by removing the power to a number divider.

**a ^{p}logb = 1/p^{a}log b**

There are several conditions for this one trait, namely: a > 0, a \ne 1.

**8. Logarithmic Principal Numbers Comparable to Numeric Powers **

The property of a base number that is proportional to the power of the numerus is a property whose numeric value is a the exponent (power) of the value of the base number which has the same result value as the value of the power of numerus that.

^{a}log a^{p }= p

There are several conditions for this one trait, namely: a > 0 and a \ne 1.

**9. Rank **

The power of logarithms is one of the properties of numbers whose powers are in the form of logarithms. The result of the power value is the value where the numerus comes from the logarithm.

**a ^{a}log m = m**

There are several conditions for this one trait, namely: a > 0, a \ne 1, m > 0.

**10. Changing the Logarithmic Base **

The nature of changing the base of this logarithm can also be broken down into a comparison of two logarithms.

^{p}log q = ^{a}log p/^{a }log q

There are several conditions for this one trait, namely: a > 0, a \ne 1, p > 0, q > 0

## Logarithmic Equation Formula

Based on the description above, logarithm is a mathematical operation which is an inverse of the exponent or power.

An example of the logarithm of the exponential form between lian: a^{b} = c if expressed in logarithmic notation it will be ^{a}logc = b.

The statement is as follows:

- a is the base or base number.
- b is the result or range of logarithms.
- c is the numerus or domain of the logarithm.

With notes:

It is necessary for you to understand, before we discuss further about the formula of the logarithm, if there is writing ^{a}log b means the same as log_{a} b.

The formula for the logarithmic equation, among others, is:

Logarithmic equation formula:If we have

^{a}logf(x) =^{a}log g(x), then f(x) = g(x) .

With some conditions such as: a > 0, a 1, f (x) > 0, g (x) > 0 .

Logarithmic inequalities:If we have log f(x) >

^{a}log g(x) then we have two states, namely:First, when a>0 means: f (x) > g (x)

Second, at time 0

## Sample Questions and Discussion

In the following, we will provide some examples of questions as well as their discussion. Listen carefully, yes.

**Sample Questions 1-3**

1. ^{2}logs 4 + ^{2}log 8 =

2. ^{2}log 32 =

3. When it is known ^{2}log 8 = m and ^{2}log 7 = n, then find the value of ^{16}logs 14!

**Answer:**

**Problem 1.**

The first step we have to do is check **the base**.

The two equations of the logarithm above, apparently have the same base value, which is 2.

Therefore, we can use the second logarithmic property to find the result.

so that, ^{2}logs 4 + ^{2}log 8 = ^{2}log (4 × 8) = ^{2}logs 32 = 5. **Remember!** The purpose of the logarithm is to find the power.

So, what 2 to the power of 32? **The answer is none other than 5**. Easy isn't it?

**Question 2.**

Let's move on to question number 2.

In question number 2, we cannot do it right away, because you will definitely experience confusion in finding the value of the power of 8 which results in 32. Then how?

If we look at the problem more closely, 8 is the result of the power of 2^{3} and also 32 which is the result of the power of 2^{5}.

Therefore, we can change the logarithmic form to:

^{8}log 32 = ^{23}log 2

= 5/3 ^{2}log 2 (use property number 6)

= 5/3(1) = 5/3

**Problem 3.**

How are you guys? Have you started to get excited yet?

*Well*, in the discussion of question number 3 this will make you even more excited!

You need to know that the model from question number 3 will often be found in National Examination questions or university selection questions *you know*.

At first glance it looks quite complicated, yes, but if you already understand the concept, this problem will be very easy to do.

If you find a problem model like this, you can find its value by using the logarithmic property of number 4.

So, the process will be:

^{2}log 8 = m and ^{2}log 7 = n, ^{16}logs 14?

^{16}log 14 = ^{2}log 14/ ^{2}log 16

**Note:**

To choose which base, we can look directly at the number that appears most often in the problem. So we know the number 2 appears 2 times, 8 as much as 1 time, and 7 as much as 1 time.

The number that appears the most is none other than 2, so we choose 2 as the basis. Got it?

= ^{2}logs (7 x 2)/ ^{2}logs (8 x 2)

Then, we **describe the numerus**.

Let's try to change it into the form already in the problem. What do you mean?

here *guys*, on the known question ^{2}log 8 and also ^{2}logs 7. Since the numbers are both 8 and 7, we break 14 into 7 × 2 and 16 into 8 × 2 so we can see the final result.

= ^{2}log 7 + ^{2}log 2/ ^{2}log 8 + ^{2}log 2 (use property number 2)

= n + 1/m + 1

**Another example question.**

**Problem 1.(EBTANAS '98)**

Is known ^{3}log 5 = x and ^{3}log 7 = y. Calculate the value of ^{3}logs 245 ^{1/2}! (EBTANAS '98)

**Answer:**

^{3}logs 245 ^{½} = ^{3}logs (5 x 49) ^{½}

^{3}logs 245 ^{½} = ^{3}logs((5) ^{½} x(49) ^{½})

^{3}logs 245 ^{½} = ^{3}logs (5) ^{½} + ^{3}logs (7^{2}) ^{½}

^{3}logs 245 ^{½} = ½( ^{3}log 5 + ^{3}logs 7)

^{3}logs 245 ^{½} = (x + y)

So, the value of ^{3}logs 245 ^{½} i.e. (x + y).

**Question 2. (UMPTN '97)**

If b = a^{4}, the values of a and b are positive, then the value of ^{a}log b – ^{b}log a ie…?

**Answer:**

It is known if b = a^{4}, then we can substitute it into the calculation to be:

^{a}log b – ^{b}loga = ^{a}log a^{4} – a^{4} log a

^{a}log b – ^{b}loga = 4 (^{a}loga) – 1/4( ^{a}logs a)

^{a}log b – ^{b}loga = 4 – 1/4

^{a}log b – ^{b}loga = 3^{3/4}

So, the value of ^{a}log b – ^{b}log a in question number 2 is 3^{3/4}.

**Problem 3. (UMPTN '97)**

If ^{a}logs (1- ^{3}log 1/27) = 2, then calculate the value of a.

**Answer:**

If we make the value 2 into a logarithm where the base number of the logarithm is a becomes ^{a}log a^{2}= 2, then we get:

^{a}logs (1- ^{3}log 1/27) = 2

^{a}logs (1- ^{3}logs 1/27) = ^{a}log a^{2}

The numeric value of the two logarithms can be an equation, namely:

1- ^{3}log 1/27 = a^{2}

^{3}logs 3 – ^{3}log 1/27 = a^{2}

^{3}logs 3 – ^{3}log 3^{(-3) }= a^{2}

^{3}logs 3/3^{-3} = a^{2}

^{3}log 3^{4} = a^{2}

4 = a^{2}

So we get the value a = 2.

**Problem 4.**

If it is known that 2log 8 = a and 2log 4 = b. Then calculate the value of 6log 14

a. 1 /2

b. (1+2) / (2+1)

c. (a+1) / (b+2)

d. (1+a) / (1+b)

**Answer:**

For 2 log 8 = a

= (log 8 / log 2) = a

= log 8 = a log 2

For 2 log 4 = b

= (log 4 / log 2) = b

= log 4 = b log 2

So ,16 log 8 = (log 16) / (log68)

= (log 2.8) / (log 2.4)

= (log 2 + log 8) / (log 2 + log 4)

= (log 2 + a log a) / (log 2 + b log b)

= log2 (1+ a) / log 2( 1+ b)

= (1+a) / (1+b)

**So, the value of 6 log 14 in the example problem above is (1+a) / (1+b). (D)**

**Question 5.**

The value of (3log 5 – 3 log 15 + 3log 9) is?

a. 2

b. 1

c. 4

d. 5

**Answer:**

(3log 5 – 3log 15 + 3log 9

= 3logs ( 5. 9) / 15

= 3log 45/15

= 3log 3

=1

**So, the value of 3log 5 – 3log 15 + 3log 9 is 1. (B)**

**Question 6.**

Calculate the value in the logarithm problem below:

- (2log 4) + (2log 8)
- (2log 2√2) + (2log 4√2)

**Answer:**

1. (2log 4 + 2log 8) = (2log 4) x 8 = 2log 3 to the power of 2 = 5

2. (2log 2√2 + 2log 4√2) = (2log 2√2) x (4√2) = 2log 16 = 4

**So, the value of each logarithm problem above is 5 and 4.**

**Question 7.**

Calculate the value in the logarithm problem below:

- 2log 5 x 5log 64
- 2 logs 25 x 5logs 3 x 3logs 32

**Answer:**

1. (2log 5) x (5log 64) = 2log 64 = 2log 26 = 6

2. (2log 25) x (5log 3) x (3log 32) =(2log 52) x (5log 3) x (3log 25)

= 2. (2log 5) x (5log 3) x 5. (3logs 2)

= 2 x 5 x (2log 5) x (5log 3) x (3log 2)

= 10 x (2log 2) = 10 x 1 = 10

**So, the value of the question above is 6 and 10.**

**Question 8.**

Calculate the value of log 25 + log 5 + log 80 is...

**Answer:**

log 25 + log 5 + log 80

= log(25 x 5 x 80)

= logs 10000

= log 104

= 4

**Problem 9.**

It is known that log 3 = 0.332 and log 2 = 0.225. Then log 18 of the question is ….

a. 0,889

b. 0,556

c. 0,677

d. 0,876

**Answer:**

Known:

- Log 3 = 0.332
- Log 2 = 0.225

Asked:

- log 18 = ….?

**Answer:**

Logs 18 = logs 9. log 2

Log 18 = (log 3.log 3). log 2

Logs 18 = 2. (0,332) + (0,225)

Log 18 = 0.664 + 0.225

Log 18 = 0.889

**So, the value of log 18 in the question above is 0.889. (A)**

**Question 10.**

Convert the following exponents into logarithmic form:

- 24 = 16
- 58 = 675
- 27 = 48

**Answer:**

*Transform the exponents into logarithmic form as follows:

If the value of ba = c, then the value for blog c = a.

- 24 = 16 → 2log 16 = 4
- 58 = 675 → 5log 675 = 8
- 27 = 48 → 2log 48 = 7

**Read also: Root Shape**

Thus a brief review this time that we can convey. Hopefully the above review can be used as your study material.