Question

How do you evaluate \[{\log _{125}}5\]?

Answer 1

Hint: We will use the concepts of algebra related to exponents and logarithms. We will define logarithms and their properties and rules. Through these definitions, we will solve this problem. And we will also look at some standard results of logarithms too.

Formula used:
Few rules or limits for a logarithm are
(1) If \[x = {\log _a}b\] then \[{a^x} = b\]. Here, \[a\] is called the base and \[b\] is called the result. But there is another term \[x\] which is called the exponent. Exponent is the power to the base which defines the number of times the base is multiplied by itself.
(2) Logarithm is not defined for a negative value i.e., \[\log n\] is not defined if \[n < 0\].
(3) Logarithm of 1 to any base is always 0. \[ {\log _a}1 = 0\], because \[{a^0} = 1\] where \[a \in R\]
(4) Logarithm to base \[e = 2.71\] is represented as \[\ln b\] which is called natural logarithm.

Complete step by step solution:
For equation \[{a^x} = b\], logarithm can be applied as \[x = {\log _a}b\]
We have to evaluate \[{\log _{125}}5\]
Let \[{\log _{125}}5 = x\]
As per logarithm definition, we can write \[{125^x} = 5\]
So, on further simplification, we can write it as, \[{\left( {{5^3}} \right)^x} = 5\]
From the formula \[{\left( {{a^m}} \right)^n} = {a^{mn}}\], we can write \[{\left( {{5^3}} \right)^x} = 5 \Rightarrow {5^{3x}} = {5^1}\]
So, here, the bases are equal, so we can equate the powers.
\[ \Rightarrow 3x = 1\]
Dividing the whole equation by 3, we get,
\[ \Rightarrow x = \dfrac{1}{3}\]
But we assumed that \[{\log _{125}}5 = x\]
Therefore, can conclude that, \[{\log _{125}}5 = \dfrac{1}{3} = 0.3333....\]

Note:
We can directly use another formula to solve this, which is \[{\log _{{b^n}}}{a^m} = \dfrac{m}{n}{\log _b}a\]
So, here, we can write this as, \[{\log _{125}}5 = {\log _{{5^3}}}{5^1} = \dfrac{1}{3}{\log _5}5 = \dfrac{1}{3}\].
There are some other standard results like \[{\log _n}\left( {ab} \right) = {\log _n}a + {\log _n}b\]
And \[{\log _n}\left( {\dfrac{a}{b}} \right) = {\log _n}a - {\log _n}b\]
Logarithm to base 10 is represented as just \[\log a\] (without representing any base)

Suppose, there is a term \[{3^5}\], which means that, 3 is multiplied by itself for five times.
So, \[{3^5} = 3 \times 3 \times 3 \times 3 \times 3\]
In the equation \[{a^x} = b\], the value of \[a\] can be found by making it the subject of the equation.
\[ \Rightarrow a = {\left( b \right)^{\dfrac{1}{x}}}\] or \[a = \sqrt[x]{b}\]
But, to find the value of \[x\], we need to apply a function called ‘Logarithm’.
So, logarithm is defined as the reverse process of exponential function.