Courses for Kids
Free study material
Offline Centres
Store Icon

The number \[{\log _2}7\] is
A.An integer
B.A rational number
C.An irrational number
D.A prime number

Last updated date: 13th Jun 2024
Total views: 384k
Views today: 8.84k
384k+ views
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 \[e\].