Question

Show that ${\log _2}7$ is an irrational number.

Answer 1

Hint: In this particular question use the concept by the assumption that the given statement is false, and use the concept that if a number is written in the form of $\dfrac{p}{q},q \ne 0$, then the number is said to be a rational number otherwise irrational number, so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Proof –
Let ${\log _2}7$ is a rational number.
So as we know that a rational number is always written in the form of ($\dfrac{p}{q},q \ne 0$), where p and q have not any common factors except 1. And p and q belong to integer values.
$ \Rightarrow {\log _2}7 = \dfrac{p}{q}$
Now as we know that ${\log _a}b = \dfrac{{\log b}}{{\log a}}$ so use this property in the above equation we have,
$ \Rightarrow \dfrac{{\log 7}}{{\log 2}} = \dfrac{p}{q}$
$ \Rightarrow q\log 7 = p\log 2$
Now as we know that $a\log b = \log {b^a}$ so use this property we have,
$ \Rightarrow \log {7^q} = \log {2^p}$
$ \Rightarrow {7^q} = {2^p}$
Now as we know that p and q belong to integer values.
So, for every integer ${2^p}$ always gives us an even number and ${7^q}$ always gives us an odd number.
So both can never be equal.
So this is a contradiction.
So our assumption is wrong.
Hence ${\log _2}7$ is an irrational number.
Hence Proved.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic properties of log such as ${\log _a}b = \dfrac{{\log b}}{{\log a}}$, $a\log b = \log {b^a}$ so apply these properties as above and simplify the given equation as above, and recall the definition of a rational number as well as the irrational number which is all stated above.