# The number \[{\log _2}7\] is

A.An integer

B.A rational number

C.An irrational number

D.A prime number

Answer

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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.

**Complete step-by-step answer:**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 \[e\].

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