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Our logarithm calculator determines the logarithmic value of a number given that you provide the log base. When the log base is provided for, you can calculate the natural log, common log and binary log values. Its just a matter of choosing. This log equation solver is a free to use tool to facilitate mathematicians, engineers, physicists and students in particular.
What is logarithm?
In Math, the log is the reverse function of exponentiation. What this implies is that the log of a given number, say x, is exponent to which another static number, the base (say b), must be elevated, to reproduce the number x.
In the simplest case, the log sums the number of occurrences of the same factor in repetitive multiplication; e.g., since \(10000 = 10 \times10 \times 10 \times 10 = 104\), the "logarithm to base 10" of 10000 is 4. The log of x to base b is represented as log b (x). This is how you can write a log.
How to calculate log with our calculator
This log base calculator is ours is exceptionally easy to use. You have to follow very simple steps to do the calculation. The user-interface of this ln calculator ensures that even this is not necessary but we will list the steps just in case.
- First, choose the log you want to calculate, for instance: natural log, common log, binary log or custom log.
- Second, input the base value.
- Third, press ‘calculate’
- That’s it. you’d get your answer
How to calculate logarithm?
Logs are very useful in measuring or calculating the large value which is beyond the reach of a simple calculator such as distance between two galaxies, growth population or volume of water on the earth. It calculates these and similar types of calculations with great accuracy.
The standard equation for log function: \(\log b(x) = y\)
In this equation, y is the logarithm, which is the number of times it is multiplied while x is the number which is given to calculate the log while b is the base which is multiplied to itself.
What is the ln function?
Ln is the natural logarithm with a base of e. e is an irrational number, which is approximately estimated as 2.718281828459. Some people confuse ln function with antilog. Do not use it as antilog because it is not an antilog. The natural logarithm is useful for calculating the time required to achieve a certain growth level.
In general, the natural logarithm of x is written as lnx, or if the basis e is implied, it is written as log x.
Product Law:
Product Law formula
\(\log_a(MN) = \log_a M + \log_a N\)
Product Law Proof
\(\math bf{Let:} \log_a M = x \Rightarrow x^a = M\) \(\mathbf{ and } N = y \Rightarrow a^y = N\)
\(\mathbf{Now} a^x \times a^y = MN \mathbf{ or } a^{x + y} = MN\)
Therefore from definition, we have,
\(\log_a(MN) = x + y = \log_a M + \log_a N\) [putting the values of x and y]
Corollary:
The law is true for more than two positive factors i.e.,
\(\log_a(MNP) = \log_a M + \log_a N + \log_a P\)
since, \(\log_a(MNP) = \log_a (MN) + \log_a P = + \log_a M + + \log_a N + \log_a P \)
Therefore in general, \(\log_a(MNP......) = \log_a M + \log_a N + \log_a P + ......\)
Quotient Law:
Quotient Law Formula
\(\log_a(\dfrac{M}{N}) = \log_a M + \log_a N\)
Quotient Law proof
\(\mathbf{Let:} \log_a M = x \Rightarrow x^a = M\) \(\mathbf{and} N = y \Rightarrow a^y = N\)
\(\mathbf{Now} a^x / a^y = M/N \mathbf{or}, a^{x - y} = M/N\)
Therefore from definition, we have,
\(\log_a(\dfrac{M}{N}) = x - y = \log_a M - \log_a N\) [putting the values of x and y]
Corollary:
\(\log_a[ \dfrac{M x \times N x \times P}{R x \times S x \times T}] \)
\( = log_a (M x \times N x \times P) - log_a (R x \times S x \times T) \)
\(\ = log_aM + log_a N + log_a P - (log_aR + log_a S + log_a T)\)
Proof of Power Rule Law:
Power Rule Formula
\(\log_aM^n = n \log_a M\)
Power Rule Proof
\(\mathbf{Let:} \log_a M = x \Rightarrow a^x = M^n\) \(\mathbf{and} M = y \Rightarrow a^y = M\)
\(\mathbf{Now} a^x = M^n = (a^y)^n = a^ny\)
\(\mathbf{Therefore} x = ny \space \mathbf{or} \space \log_a M^n = n \log_a M \) [putting the values of x and y]
Proof of Change of base Rule Law:
Change of base Law formula
\(\log_a M = \log_b M \times \log_b \)
Change of base Law proof
\(\mathbf{Let:} \log_a M = x \Rightarrow a^x = M^n\) ,
\( M = y \Rightarrow b^y = M\)
\(\mathbf{and} \log_a b = z \Rightarrow a^z = b\)
\(\mathbf{Now} a^x = M = b^y - (a^z)y = a^yz\)
\(\mathbf{Therefore} x = yz \space \mathbf{or} \space \log_a M = \log_b M \times \log_a b \) [putting the values of x ,y and z]
Corollary:
(i) Putting M = a on both sides of the change of base rule formula [\(\log_a M = \log_b M \times \log_b \)] we get,
\(\log_a a = \log_b a \times \log_a b = 1\) [since, loga a = 1]
\(\log_b a = \dfrac{1}{\log_a b}\)
\(\log_b M = \dfrac{\log_a M}{\log_a b} \)