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Logarithms Formulas | Basic & Special Case Log with Examples Equations

Log Formulas

Logarithm is a convenient way to express large numbers. The logarithm of a product is the sum of the logarithms of the numbers being multiplied. Log Formulas are available below for the convenience of the students. Here you can find Logarithms Formulas including Basic & Special Case Log with Examples Equations that can help you to be familiar with this topic easily. Apart from it, you may check Logarithm Numerical also that can help you to know about the way of applying Log Formulas, so please have a look..

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Log Formulas

Basic idea and rules for logarithms

A logarithm is the opposite of a power. In other words, if we take a logarithm of a number, we undo an exponentiation

Let’s start with simple example. If we take the base b=2 and raise it to the power of k=3, we have the expression 23. The result is some number, we’ll call it cc, defined by 23=c, We can use the rules of exponentiation to calculate that the result is

c=23=8.

Let’s say I didn’t tell you what the exponent k was. Instead, I told that the base was b=2 and the final result of the exponentiation was c=8. To calculate the exponent k, you need to solve

2k=8.

From the above calculation, we already know that k=3. But, what if I changed my mind, and told you that the result of the exponentiation was c=4 =, so you need to solve 2k=4? Or, I could have said the result was c=16  (solve 2k=16) or c=1 (solve 2k=1).

A logarithm is a function that does all this work for you. We define one type of logarithm (called “log base 2” and denoted log2) to be the solution to the problems I just asked. Log base 2 is defined so that

log2c=k

is the solution to the problem

2k=c

for any given number cc. In other words, the logarithm gives the exponent as the output if you give it the exponentiation result as the input. To get all answers for the above problems, we just need to give the logarithm the exponentiation result cc and it will give the right exponent k of 2. The solution to the above problems are:

Just like we can change the base bb for the exponential function, we can also change the base bb for the logarithmic function. The logarithm with base bb is defined so that

logbc=k

is the solution to the problem

bk=c

For any given number c and any base b.

For example, since we can calculate that 103=1000, we know that log101000= 3 (“log base 10 of 1000 is 3”). Using base 10 is fairly common. But, since in science, we typically use exponents with base e, it’s even more natural to use e for the base of the logarithm. This natural logarithm is frequently denoted by ln⁡(x), i.e.,

ln(x)=logex.

In other words,

k=ln(c)……………(1)

is the solution to the problem

ek=c ………….(2)

for any number c. Since using base e is so natural to mathematicians, they will sometimes just use the notation log x instead of ln x. However, others might use the notation logx for a logarithm base 10, i.e., as a shorthand notation for log10x. Because of this ambiguity, if someone uses log x without stating the base of the logarithm, you might not know what base they are implying. In that case, it’s good to ask.

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Basic Log Formulas

Since taking a logarithm is the opposite of exponentiation (more precisely, the logarithmic function logbx  is the inverse function of the exponential function bx), we can derive the basic rules for logarithms from the basic rules for exponents.

For simplicity, we’ll write the rules in terms of the natural logarithm ln(x). The rules apply for any logarithm logbx, except that you have to replace any occurence of e with the new base bb.

The natural log was defined by equations (1) and (2). If we plug the value of k from equation (1) into equation (2), we determine that a relationship between the natural log and the exponential function is

elnc=c …………………….(3)

Or, if we plug in the value of cc from (2) into equation (1), we’ll obtain another relationship

ln(ek)=k    ……………………..4

These equations simply state that ex and ln x are inverse functions. We’ll use equations (3) and (4) to derive the following rules for the logarithm..

All Log Formulas (BASIC)

Rule or special caseFormula
Productln(xy)=ln(x)+ln(y)
Quotientln(x/y)=ln(x)−ln(y)
Log of powerln(xy)=yln(x)
Log of eeln(e)=1
Log of oneln(1)=0
Log reciprocalln(1/x)=−ln(x)

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Formulas Of Log – The product rule

The quotient rule

The quotient rule for logarithms follows from the quotient rule for exponentiation

ea/eb= e a−b

in the same way.

Starting with c=x/y in equation (3) and applying it again with c=x and c=y, we can calculate that

eln(x/y) = x/y

=e ln(x) / e ln(y) = e ln(x)−ln(y),

Where in the last step we used the quotient rule for exponentation with a=ln(x)a=ln⁡(x) and b=ln(y)b=ln⁡(y). Since eln(x/y)=e ln(x)−ln(y) we can conclude that the quotient rule for logarithms is

ln(x/y)=ln(x)−ln(y)

(This last step could follow from, for example, taking logarithms of both sides of eln(x/y)=eln(x)−ln(y) like we did in the last step for the product rule.)

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Log of a power

To obtain the rule for the log of a power, we start with the rule for power of a power,

(ea)b=eab

Starting with c=xy in equation  (3) and applying it again, this time just once more with c=x, we can calculate that

e ln(xy)=xy

= (eln(x))y

=e yln(x)

Where in the last step we used the power of a power rule for a=ln(x) and b=y From eln(xy)=eyln(x), we can conclude that

ln(xy)=yln(x)

which is the rule for the log of a power.

Log of e

The formula for the log of e comes from the formula for the power of one,

e1=e

Just take the logarithm of both sides of this equation and use equation (4) to conclude that

ln(e)=1

Log of one

The formula for the log of one comes from the formula for the power of zero

e0=1

Just take the logarithm of both sides of this equation and use equation (4)(4) to conclude that

ln(1)=0.

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Log of reciprocal

The rule for the log of a reciprocal follows from the rule for the power of negative one

x−1=1/x

and the above rule for the log of a power. Just substitute y=−1y=−1 into the the log of power rule, and you have that

ln(1/x)=−ln(x).

Question and Answers Based On Logarithms Formulas

Question1. Which of the following statements is not correct?

  1. log10 10 = 1
  2. log (2 + 3) = log (2 x 3)
  3. log10 1 = 0
  4. log (1 + 2 + 3) = log 1 + log 2 + log 3

Answer: Option 2

Question2. If log 2 = 0.3010 and log 3 = 0.4771, the value of log5 512 is:

  1. 870
  2. 967
  3. 876
  4. 912

Answer: Option 3

Question3. log 8/ log 8 is equal to:

  1. 1/8
  2. 1/4
  3. 1/2
  4. 1/8

Answer: Option 3

Question4.  If log 27 = 1.431, then the value of log 9 is:

  1. 934
  2. 945
  3. 954
  4. 958

Answer: Option 3

Question5. If log a/b + log b/a = log (a + b), then:

  1. a + b = 1
  2. a – b = 1
  3. a = b
  4. a2 – b2 = 1

Answer: Option 1

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Logarithms Formulas

In mathematics there are different types of formulas and algorithms. One of them is Log Formulas. Logarithms Formulas are used in mathematics while solving trigonometric equations etc. logarithm is the opposite function to exponentiation, just as division is the inverse of multiplication and vice versa.

Here on this page we are going to provide Basic & Special Case Log Product/Power Rule Equations of Logarithms. Candidates willing to learn Log Formulas easily must take a look below.

Here on this page of recruitmentresult.com we are going to provide all the Logarithms Formulas – Basic & Special Case Log Product/Power Rule Equations that are used by candidates while solving mathematical equations.

Use of Log starts from 11th standards and continue till graduation/ Post-graduation in courses like BCA / MCA / B Tech / BE / M Tech / ME etc. for the ease of students here we have also given All Log Formulas PDF.

Final Words!!

We hope the information given here on this page will prove to be useful for all of you regarding the topic Integration Log Formulas.

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