Logarithms

πŸ†Practice rules of logarithms

What Are Logarithmic Laws?

There are a few logarithmic laws worth knowing to make solving problems easier. The following laws are the main rules you will use. It should be noted that the letters a, m, n must be positive real numbers for these laws to be valid.

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Test yourself on rules of logarithms!

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\( \log_{10}3+\log_{10}4= \)

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These are the rules

Constant Values:

It can be automatically determined that:

  • loga(1)=0 log_a\left(1\right)=0
  • loga(a)=1 log_a\left(a\right)=1

Basic Arithmetic Operations

Multiplication, division, subtraction, and addition operations between logarithms:

  • logaMN=logaM+logaN log_aMN=log_aM+log_aN
  • logaM/N=logaMβˆ’logaN log_aM/N=log_aM-log_aN
  • Loga(M)Γ—Logn(D)=Logn(M)Γ—Loga(D) Log_a\left(M\right)\times Log_n\left(D\right)=Log_n\left(M\right)\times Log_a\left(D\right)
  • LogaMn=nLogaM Log_aM^n=nLog_aM

Changing the Base of a Logarithm:

It should be noted that on calculators the default value is a logarithm based on 10 10 . But sometimes we want to calculate a logarithm with a base other than 10 10 . For example, what if we want to know to what power we must raise 2 2 to get the number 4 4 ? (The answer is, of course, 2 2 , because 2 2 raised to the power of 2 2 equals 4 4 ). To perform this change, the base change of the algorithm must be used. There are two ways to do this:

  • logb(x)=logc(x)/logc(b) log_b\left(x\right)=log_c\left(x\right)/log_c\left(b\right)
  • logb(c)=1/logc(b) log_b\left(c\right)=1/log_c\left(b\right)

Derivative of the Logarithm:

fx=logb(x)β‡’fβ€²x=1/xln(b) fx=log_b\left(x\right)β‡’f^{\prime}x=1/xln(b)


Integral of the Logarithm:

∫logb(x)dx=xΓ—logb(x)βˆ’1/ln(b)+C ∫log_b\left(x\right)dx=x\times log_b\left(x\right)-1/ln\left(b\right)+C



Methods for Calculating Logarithms

There's a main method for calculating logarithms in high school, based on the definition of the log term. As mentioned, log is actually an inverse operation to a normal power. Therefore, if we want to find out what log⁑39 \log_39 is, we have to ask ourselves, "What power of three would be equal to nine?" And we'll find that the answer is two. In other words, we can mark the answer with an X X and create an equation:

log⁑39=X \log_39=X

We can say that: 3X=9 3^X=9

And create a simple exponential equation for the solution.

In more complicated exercises, previous logarithmic laws should be used to simplify the solution.

For example: take the logarithm

log⁑X125=3 \log_X125=3

According to the Logarithmic Law we can say that:

X3=125 X^3=125 And so we can know that X=5 X=5

With a bit of practice, you'll see that solving logarithms isn't as scary or daunting as it seems, and like everything in math, practice and understanding are the name of the game!


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Exponential Functions

First, to understand what logarithms are, one must understand what an exponential function is. An exponential function is one of the most useful functions in the language of mathematics and expresses interesting growth and decay processes. Due to the great importance of the exponential function, it is one of the important topics in the study of mathematics. An exponential function is actually a function of the type aXa ^ X, where a is the base of the function.

A well-known exponential function is one based on the number e, which is a mathematical constant equal to the value 2.718282.71828. This function is special for several reasons, one of which is that its derivative is equal to itself.

A logarithmic function is the opposite of an exponential function, and it actually answers the question: "To what power do we need to raise a given number to obtain another given number?" For example, if I want to know to what power I need to raise 1010 to get the number 100100, on the calculator, I would enter: log100100. The result will be 22, since 1010 raised to the power of 22 is 100100.


logarithm foto

Natural Logarithm

So far we have only seen what a common logarithm is, but there is another type of logarithm, which is known as the natural logarithm, and this logarithm is the inverse function of the exponential. That is, the natural logarithm has a base e e , this base is an irrational number which is a constant and its value is:

e=2.71828182… e=2.71828182\ldots

And the natural logarithm can be represented as:

ln⁑x=y \ln x=y

Properties of the natural logarithm

  • ln⁑1=0 \ln1=0

Although the logarithm does not have a base as such, it is understood that its base is e e , so we must raise e e to the power of zero to get the result 1 1 .

e0=1 e^0=1

  • ln⁑e=1 \ln e=1

We can see this as: e1=e e^1=e

  • ln⁑ex=x \ln e^x=x
  • eln⁑x=x e^{\ln x}=x

These two properties are because they are inverse operations.


If you're interested in this article, you might also be interested in the following articles:

  • Powers
  • The Rules of Exponentiation
  • Division of Powers with the Same Base
  • Power of a Quotient
  • Power of a Power
  • Power with Zero Exponent
  • Powers with a Negative Integer Exponent
  • Taking Advantage of All the Properties of Powers or Laws of Exponents
  • Exponentiation of Whole Numbers

On the Tutorela blog, you'll find a variety of articles about mathematics.


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Cheat Sheets, Logarithmic Rules

Hojas de trucos reglas logarΓ­tmicas

Logarithmic Laws Exercises

Exercise 1

Assignment

40 40% is equal to 2 2 . What is the whole number?

Solution

To find the whole number, we multiply a fractional number by 100 100 and divide by the percentage number percentages

100β‹…240%=20040=204=5 \frac{100\cdot2}{40\%}=\frac{200}{40}=\frac{20}{4}=5

Answer

5 5


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Exercise 2

Assignment

3log⁑49+8log⁑413= 3\log_49+8\log_4\frac{1}{3}=

Solution

Where:

3log⁑49=log⁑493=log⁑4729 3\log_49=\log_49^3=\log_4729

and

8log⁑413=log⁑4(13)8= 8\log_4\frac{1}{3}=\log_4\left(\frac{1}{3}\right)^8=

log⁑4138=log⁑416561 \log_4\frac{1}{3^8}=\log_4\frac{1}{6561}

Therefore

3log⁑49+8log⁑413= 3\log_49+8\log_4\frac{1}{3}=

log⁑4729+log⁑416561 \log_4729+\log_4\frac{1}{6561}

log⁑ax+log⁑ay=log⁑axy \log_ax+\log_ay=\log_axy

(729β‹…16561)=log⁑419 \left(729\cdot\frac{1}{6561}\right)=\log_4\frac{1}{9}

log⁑49βˆ’1=βˆ’log⁑49 \log_49^{-1}=-\log_49

Answer

βˆ’log⁑49 -\log_49


Exercise 3

Task

2log⁑82+log⁑83= 2\log_82+\log_83=

Solution

2log⁑82=log⁑822=log⁑84 2\log_82=\log_82^2=\log_84

2log⁑82+log⁑83=log⁑84+log⁑83= 2\log_82+\log_83=\log_84+\log_83=

log⁑84β‹…3=log⁑812 \log_84\cdot3=\log_812

Answer

log⁑812 \log_812


Do you think you will be able to solve it?

Exercise 4

Assignment

12log⁑24Γ—log⁑38+log⁑39Γ—log⁑37= \frac{1}{2}\log_24\times\log_38+\log_39\times\log_37=

Solution

Let's break it down into parts

log⁑24=x \log_24=x

2x=4 2^x=4

x=2 x=2

log⁑39=x \log_39=x

3x=9 3^x=9

x=2 x=2

We substitute back into the equation

12β‹…2log⁑38+2log⁑37= \frac{1}{2}\cdot2\log_38+2\log_37=

1β‹…log⁑38+2log⁑37= 1\cdot\log_38+2\log_37=

log⁑38+log⁑372= \log_38+\log_37^2=

log⁑38+log⁑349= \log_38+\log_349=

log⁑3(8β‹…49)=log⁑3392 \log_3\left(8\cdot49\right)=\log_3392

Answer

log⁑3392 \log_3392


Exercise 5

Assignment

14β‹…log⁑61296β‹…log⁑612βˆ’log⁑63= \frac{1}{4}\cdot\log_61296\cdot\log_6\frac{1}{2}-\log_63=

Solution

Let's break it down into parts

log⁑61296=x \log_61296=x

6x=1296 6^x=1296

x=4 x=4

14β‹…4β‹…log⁑612βˆ’log⁑63= \frac{1}{4}\cdot4\cdot\log_6\frac{1}{2}-\log_63=

log⁑612βˆ’log⁑63= \log_6\frac{1}{2}-\log_63=

log⁑6(12:3)=log⁑616 \log_6\left(\frac{1}{2}:3\right)=\log_6\frac{1}{6}

log⁑616=x \log_6\frac{1}{6}=x

6x=16 6^x=\frac{1}{6}

x=βˆ’1 x=-1

Answer

βˆ’1 -1


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Exercise 6

Assignment

log⁑7x4βˆ’log⁑72x2=3 \log_7x^4-\log_72x^2=3

Find X X

Solution

log⁑axβˆ’log⁑ay=log⁑axy \log_ax-\log_ay=\log_a\frac{x}{y}

log⁑7x4βˆ’log⁑72x2= \log_7x^4-\log_72x^2=

log⁑7x42x2=3 \log_7\frac{x^4}{2x^2}=3

73=x22 7^3=\frac{x^2}{2}

Multiply by: 2 2

2β‹…73=x2 2\cdot7^3=x^2

Extract the square root

x=680=714 x=\sqrt{680}=7\sqrt{14}

x=βˆ’680=βˆ’714 x=-\sqrt{680}=-7\sqrt{14}

Answer

βˆ’714Β Β ,Β 714 -7\sqrt{14\text{ }}\text{ , }7\sqrt{14}


Exercise 7

Assignment

log⁑7x+log⁑(x+1)βˆ’log⁑7=log⁑2xβˆ’log⁑x \log7x+\log(x+1)-\log7=\log2x-\log x

How much is X X ?

Solution

Definition domain

x>0 x>0

x+1>0 x+1>0

x>βˆ’1 x>-1

log⁑7x+log⁑(x+1)βˆ’log⁑7=log⁑2xβˆ’log⁑x \log7x+\log\left(x+1\right)-\log7=\log2x-\log x

log⁑7xβ‹…(x+1)7=log⁑2xx \log\frac{7x\cdot\left(x+1\right)}{7}=\log\frac{2x}{x}

We reduce by: 7 7 and by X X

x(x+1)=2 x\left(x+1\right)=2

x2+xβˆ’2=0 x^2+x-2=0

(x+2)(xβˆ’1)=0 \left(x+2\right)\left(x-1\right)=0

x+2=0 x+2=0

x=βˆ’2 x=-2

Not in the definition domain x>0 x>0

xβˆ’1=0 x-1=0

x=1 x=1

Definition domain

Answer

1 1


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Review Questions

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What are the elements of a logarithm?

The logarithm function has four elements:

log⁑ax=b \log_ax=b

Where:

log⁑ \log is its symbol

x x is the argument of the logarithm

a a is the base of the logarithm

b b is the result of the logarithm

It reads: "The logarithm base a a of x x equals b b ”


What is the relationship between the exponential function and the logarithmic function and some examples?

A logarithm is the inverse operation of the exponential function, it is important to remember that all operations have an inverse operation, in this case, every logarithmic function can be related to an exponential one.

Example 1:

Calculate

log⁑416= \log_416=

We must find a number which will be the exponent that the base 4 4 must have to obtain the logarithm's argument, which is 1616, in this case the result will be 2 2 , since

42=16 4^2=16

Result

2 2

Example 2:

Calculate

log⁑8512= \log_8512=

Now we must find the exponent that the base 8 8 must have to obtain as a result the argument of the logarithm, in this case the result is 3 3 , because:

83=512 8^3=512

Result

3 3

From these examples, we can conclude the very close relationship that exists between the logarithmic function and the exponential function.


What is a logarithm used for?

A logarithm is used to determine what number the base of the logarithm must be raised to in order to obtain the argument.


What is a logarithm and examples?

As we have already said, a logarithm is the inverse operation of an exponential. Therefore, the logarithmic function is the one where we must calculate the exponent's number that the base must have so that it gives us the number of the argument; here we will present some examples:

Examples:

  • log⁑72401=4 \log_72401=4

Since, 74=2401 7^4=2401

  • log⁑232=5 \log_232=5

As 25=32 2^5=32

  • log⁑9729=3 \log_9729=3

This is because: 93=729 9^3=729

  • log⁑100=2 \log_{}100=2

In this last example, the logarithm does not have an explicit base, but when a logarithm does not have a base as such, we must assume that it is a base 10 10 logarithm, therefore the answer to this logarithm is 2 2 , since 102=100 10^2=100 .


What are the laws of logarithms, examples?

Let's look at some of the laws of logarithms:

  • log⁑a1=0 \log_a1=0
  • log⁑aa=1 \log_aa=1
  • alog⁑ax=x a^{\log_ax}=x , this being because they are inverse functions.
  • log⁑am+log⁑an=log⁑a(mΓ—n) \log_am+\log_an=\log_a\left(m\times n\right)
  • log⁑amβˆ’log⁑an=log⁑a(mn) \log_am-\log_an=\log_a\left(\frac{m}{n}\right)
  • log⁑axr=rlog⁑ax \log_ax^r=r\log_ax

It should be mentioned that there are no logarithms of negative numbers, nor of negative bases, and neither does the logarithm of zero exist.

Examples where we can apply these laws

Example 1:

log⁑88=1 \log_88=1

And this law is too obvious since 81=8 8^1=8 , therefore the exponent must be 1 1 , to obtain the argument of the logarithmic function. And that is why law 2 2 mentioned above means that if you have a logarithm with the same base and argument, the result will always be 1 1 .

Example 2:

Find the value of x x , in the following logarithmic equation

log⁑3x+log⁑327=5 \log_3x+\log_327=5

We can use the sum of logarithms and obtain:

log⁑327x=5 \log_327x=5

We use the third law mentioned above:

3log⁑327x=35 3^{\log_327x}=3^5

We have:

27x=243 27x=243

From here we solve for x x

x=24327=9 x=\frac{243}{27}=9

Therefore, the value of x=9 x=9

And we can verify:

log⁑3x+log⁑327=5 \log_3x+\log_327=5 , here we substitute the value we have found for x=9 x=9

log⁑39+log⁑327=5 \log_39+\log_327=5 and as we well know:

log⁑39=2 \log_39=2 and

log⁑327=3 \log_327=3 , then;

log⁑3x+log⁑327=2+3=5 \log_3x+\log_327=2+3=5

Result

x=9 x=9

Another way to solve this exercise is as follows:

log⁑3x+log⁑327=5 \log_3x+\log_327=5

From here we can calculate the second logarithm:

log⁑327=3 \log_327=3 , then we substitute this value

(log⁑3x)+3=5 \left(\log_3x\right)+3=5

From here we solve for the logarithm:

log⁑3x=5βˆ’3 \log_3x=5-3

log⁑3x=2 \log_3x=2

Then here we already have the exponent that the base 3 3 must have, which should be 2 2 , therefore 32=9 3^2=9 , and from here we obtain the argument of the algorithm, that is the value of x x

Or using law 3 3 we have

3log⁑3x=32 3^{\log_3x}=3^2

x=9 x=9

Result

x=9 x=9


What is the difference between a common logarithm and a natural logarithm?

The difference is only the base, as the common logarithm has base 10 10 : log⁑10 \log_{10} and the natural logarithm has base e e

ln⁑e \ln_e

Although in reality both logarithms do not have their bases implicit, each of the bases should be understood.


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