Value of Log 2

Exponentiation to Logarithm

The formal definition of logarithms is that it is the inverse of exponentiation. This is very important to often make the plots of data easier to understand. To assess the quality of sound, the Richter scale to measure the impact of an earthquake, etc we use a logarithmic scale. The two most important logarithms we use are natural logarithms or the logarithms to the base ‘e’ and logarithm to the base 10. Natural logarithms are often written ln(x) whereas log(x) often means logarithm to the base is 10.


Value of Log

Let us try to understand how to do logarithmic operations using the example log2. An exponential is written of the form xn = m. This means if we multiply x, n times we get m. For example 23 = 8. 


Now if we take the logarithm of m to the base x we get n. That means n is the exponential of x to obtain m. This is written as follows: Logx(m) = n. So taking the previous example, we get log2(8) = 3. The log value of a number depends on the base.


Some Important Properties of Logarithm

The following properties of logarithm help us to solve problems easier.

  • Logb(b) = 1.

This property is very easy to understand. We know if bn = b, then n should be 1. Hence, the property.

  • logb(MN) = logb(M) + logb(N) .

This is called the product rule. Let us try to understand Why this rule works. Suppose bm = M and bn = N. Then, b(m+n) = MN.

Taking logarithm of MN, we get logb(MN) = m+n (from the definition of logarithm). Now logb(M) = m and logb(N) = n.

That is, logb(MN) = logb(M) + logb(N)

  • Logb(am) = m*logb(a)

This is known as the law of exponents.

Logb(am) = logb(a*a*a*….(m times)).From the product rule we know this is equal to

Logb(a) + logb(a) +……. m times = m*logb(a)

  • Logb(M/N) = logb(M) – logb(N)

  • Logb(m) = loga(m)/loga(b). This is known as the change – of –base formula. 

  • The logarithm of 0 or negative numbers is not defined. 

  • The logarithm of a number between 0 and 1 gives a negative value.

  • The logarithm of 1 is 0.


Value of Log(2)

As stated earlier to find the value of the log of any number we should know the base. Here it is always easier to find the value of log(2) to the base 10 and use it to find the value of the logarithm of 2 to any other bases. Using properties we can find the answer through calculations, close to the actual value.


We know 210 is 1024. Taking the logarithm, we get

 10 *log(2) = log(1024).


Approximating 1024 to 1000, 

Log(2) = \[\frac{log(1000)}{10}\] = \[\frac{log(10^{3})}{1}\]. From exponential rule, we can write this as \[\frac{3log(10)}{10}\].


Since the base is also 10, we get log(2) = 3*0.1. = 0.3. This is a very accurate value as the value we obtain using a calculator is 0.301. 


We can use the expansion formula of the natural logarithm to find the value of ln(2).

ln x = Substituting the value 2 in place of x, we obtain the value of ln(x) = 0.69.

Using these values we can find the value of other logarithms too.

We have discussed the basic meaning of logarithms, their important properties, and formulas and how to find values of logs of numbers in brief.

FAQ (Frequently Asked Questions)

1.Prove the change of base formula of a logarithm.

The change of base formula of logarithm states that logb(m) = loga(m)/loga(b).

Let bs = m, ar = m and at = b. 


Now, suppose logb(m) = y; Taking antilog we get m= by.


Taking log on both sides to the base of a, we get,


loga(m) = y*loga (b).(From law of exponents we know Logb(am) = m*logb(a))


Dividing both sides by loga(b) , we get 


loga(m)/ loga(b) = y* loga (b)/ loga(b)


That is, loga(m)/ loga(b) = y.


Substituting the value of y we get, logb(m) = loga(m)/loga(b).

2.Find the values of  

  1. Log(23) and ln(23)

  2. Log(400)

  3. Log4(2)

Answer

1.From the law of exponents, we know that log(ab) = b*log(a). Therefore, log(23) = 3*log(2). We know, log(2) = 0.3. Therefore,


log(23) =3*0.3 = 0.9.


Similarly, we know ln(2) = 0.69. That is,


ln(23) = 3* 0.69 = 2.07.


2.Log(400) = log(4*100).

From the product rule, we know that this is equal to log(4) + log(100).


Log(4) = 2* log(2) = 2*0.3 = 0.6.


Log(100) = log(102) =2* log(10) = 2 (logb(b)=1).


Therefore, log(400) = 2.6.


3.From the formula of change of bases we know,

Logb(m) = loga(m)/loga(b).


That is log4(2) = log10(2)/log10(4).


We know log10(2) = 0.3 and log10(4) = 0.6


That is, log4(2) = 0.5