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# The number ${\log _2}7$ isA.An integerB.A rational numberC.An irrational numberD.A prime number  Verified
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Hint: We have given ${\log _2}7$. We have to tell whether ${\log _2}7$ is rational, irrational, an integer or a prime number. Firstly, we consider that ${\log _2}7$ is a rational number and solve it for a rational number. If there is no contradiction, then it will be rational otherwise it will be an irrational number.

We have given that ${\log _2}7$.
We have to find its value whether it is rational or irrational or prime number or an integer.
Let us consider that ${\log _2}7$ is a rational number, then there exists an integer $p$ and $q$ such that${\log _2}7 = \dfrac{p}{q}$, where $q$ is not equal to zero.
So, ${2^{\dfrac{p}{q}}} = 7$
Taking exponent ‘q’ on both sides, we get:
${2^p} = {7^q}$
Now, I know that ${2^p}$ is an even number for all values of $p$ and ${7^q}$ is an odd number for all values of $q$. Therefore, there is no integer for which ${2^p} = {7^q}$
So, our supposition is wrong.
${\log _2}7$ is not a rational number.
We know that every integer is a rational number and ${\log _2}7$ is not rational, so ${\log _2}7$ is also not an integer. It will also not be the prime number as every prime number is integer.
So, ${\log _2}7$ is an irrational number.
Option (C) is correct.

Note: Logarithmic function is inverse of exponential function. The logarithmic function is defined as for $x > 0,a > 0$ and $a \ne 1,y = {\log _a}x$ if and only if $x = {a^y}$, then the function is given as $f\left( x \right) = \log {}_ax$.
The base of logarithm is $a$. This can be read as $\log$ base $a$ of $x$. The most common bases used in logarithmic functions are base $10$ and base $e$.