Question

How to find $x$ for ${{\log }_{5}}x=4?$

Answer 1

We will use the law of logarithms to find the value of the variable $x.$ That is, to solve this problem we will use the law ${{\log }_{b}}x=n\Leftrightarrow x={{b}^{n}}.$ We will use the identity given by ${{b}^{n}}={{\left( {{b}^{p}} \right)}^{q}}$ if $n=p\cdot q.$

Complete step-by-step solution:
Consider the given equation ${{\log }_{b}}x=4.$
This is read as log $x$ to the base $b$ equals to $4.$
We are asked to find the value of $x$ for which the value of the base $b$ logarithm is $4.$
So, here, we are going to us the law of logarithms given as ${{\log }_{b}}x=n\Leftrightarrow x={{b}^{n}}.$
In this equation, $b$ is the base value as we have said earlier and $n$ is any natural number that is the value of the given base $b$ logarithm.
This law is explained below:
If we are given with a base $b$ logarithm of a variable $x$ that produces a value $n,$ then the value of the variable is equal to the base $b$ to the power $n.$ The converse is also true. That is, if the value of a variable $x$ is equal to a number $b$ to the power $n,$ then the logarithm of $x$ to the base $b$ is equal to the number $n.$
Let us apply the law in our problem.
If we are comparing the values, we will get to know that $b=5$ an $n=4$ in the concerned problem.
So, we can apply the values to get the solution of the equation ${{\log }_{5}}x=4.$
Therefore, ${{\log }_{5}}x=4\Leftrightarrow x={{5}^{4}}.$
We have the identity ${{b}^{n}}={{\left( {{b}^{p}} \right)}^{q}}$ if $n=p\cdot q.$
So, we will get ${{5}^{4}}={{\left( {{5}^{2}} \right)}^{2}}$ and ${{5}^{2}}=25.$ This will give us ${{25}^{2}}=625.$
Therefore, $x=625.$
Hence the value of $x=625$ for which ${{\log }_{5}}x=4.$

Note: The logarithm with base $10$ is called the common logarithm. The logarithm with base $e,$ is called the natural logarithm. It is represented as $\ln .$ Natural logarithm is the inverse of the exponential function.