The number \[{\log _2}7\] is
1) Integer
2) Rational number
3) Irrational number
4) Prime number
Answer
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Hint: Here in this question, we have to determine the value of the logarithm function by considering the logarithm property and then we have to say that the resultant value belongs to integer or rational or irrational or prime numbers depending upon the nature of the logarithmic value.
Complete step-by-step answer:
The function from positive real numbers to real numbers to real numbers is defined as \[{\log _b}:{R^ + } \to R \Rightarrow {\log _b}\left( x \right) = y\], if \[{b^y} = x\], is called logarithmic function or the logarithm function is the inverse form of exponential function.
Consider the given logarithm function:
\[{\log _2}7\]
Here, mentioned logarithm function having base value 2.
Let us assume \[{\log _2}7\] be a rational number which is in the form of \[\dfrac{p}{q}\], then
\[{\log _2}7 = \dfrac{p}{q}\], where \[p\], \[q \in I\] and \[q \ne 0\].
\[ \Rightarrow {\log _2}7 = \dfrac{p}{q}\]
Take a antilog with base 2 on both side, then we have
\[ \Rightarrow 7 = {2^{\dfrac{p}{q}}}\]
Take \[{q^{th}}\] power on both side, then we have
\[ \Rightarrow {\left( 7 \right)^q} = {\left( {{2^{\dfrac{p}{q}}}} \right)^q}\]
Apply the property of exponent i.e., \[{\left( {{a^m}} \right)^n} = {a^{mn}}\], then
\[ \Rightarrow {7^q} = {2^p}\]
Since, \[{2^p}\] is even for all integers and \[{7^q}\] is odd for all integers and no integer can be both even and odd.
Hence, our assumption is wrong, that is \[{\log _2}7\] not a rational number.
Therefore, by contradiction \[{\log _2}7\] is an irrational number.
Hence the option 3) is a rational number.
So, the correct answer is “Option B”.
Note: Before solving the questions, first need to know some definitions of number system i.e.,
Rational Numbers- Numbers that can be written in the form of \[\dfrac{p}{q}\] where \[q \ne 0\].
Irrational Numbers- All the numbers which are not rational and cannot be written in the form of \[\dfrac{p}{q}\].
Integers - the numbers which can be positive, negative or zero, but cannot be a fraction.
Prime numbers – the positive integers having only two factors, 1 and the integer itself.
Complete step-by-step answer:
The function from positive real numbers to real numbers to real numbers is defined as \[{\log _b}:{R^ + } \to R \Rightarrow {\log _b}\left( x \right) = y\], if \[{b^y} = x\], is called logarithmic function or the logarithm function is the inverse form of exponential function.
Consider the given logarithm function:
\[{\log _2}7\]
Here, mentioned logarithm function having base value 2.
Let us assume \[{\log _2}7\] be a rational number which is in the form of \[\dfrac{p}{q}\], then
\[{\log _2}7 = \dfrac{p}{q}\], where \[p\], \[q \in I\] and \[q \ne 0\].
\[ \Rightarrow {\log _2}7 = \dfrac{p}{q}\]
Take a antilog with base 2 on both side, then we have
\[ \Rightarrow 7 = {2^{\dfrac{p}{q}}}\]
Take \[{q^{th}}\] power on both side, then we have
\[ \Rightarrow {\left( 7 \right)^q} = {\left( {{2^{\dfrac{p}{q}}}} \right)^q}\]
Apply the property of exponent i.e., \[{\left( {{a^m}} \right)^n} = {a^{mn}}\], then
\[ \Rightarrow {7^q} = {2^p}\]
Since, \[{2^p}\] is even for all integers and \[{7^q}\] is odd for all integers and no integer can be both even and odd.
Hence, our assumption is wrong, that is \[{\log _2}7\] not a rational number.
Therefore, by contradiction \[{\log _2}7\] is an irrational number.
Hence the option 3) is a rational number.
So, the correct answer is “Option B”.
Note: Before solving the questions, first need to know some definitions of number system i.e.,
Rational Numbers- Numbers that can be written in the form of \[\dfrac{p}{q}\] where \[q \ne 0\].
Irrational Numbers- All the numbers which are not rational and cannot be written in the form of \[\dfrac{p}{q}\].
Integers - the numbers which can be positive, negative or zero, but cannot be a fraction.
Prime numbers – the positive integers having only two factors, 1 and the integer itself.
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