If $\log_27=x$ is $x$ irrational or rational?

Question

$x$ will definitely be between $2$ and $3$, but why does it have to irrational? Let’s assume that you don’t have the log table and no calculator, then how do you determine if it’s rational or irrational?

It’s probably an obvious answer like $x$ will always be irrational and all that, but I’d still like to know.

Thanks a lot.

Answer 1

Hint $\log_2 (7) =x \Leftrightarrow 2^x=7$. Now, if $x=\frac{m}{n}$ would be rational then $$2^m=7^n$$

Answer 2

Here is a proof by contradiction:

Suppose $x=p/q$, to be rational. We have $2^{p}=7^q$. But the right side is odd whole the left side is even. Thus, we have a contradiction. So $x$ is irrational.

Answer 3

Another approach:

$\log _2 a=x$ is integer if and only if $a=2^n; n∈N$

But $7=2^3-1$, so x is not integer which means it is a fraction or a rational number with denominator greater than unity.

Answer 4

For $\text{log}_{2}(7)$ to be rational, then $2^\frac{a}{b} = 7$, and thus $2^{a} = 7^{b}$ for some integers $a$ and $b$. However, $7^{b}$ will always be odd, since $7$ is odd, and $2^{a}$ is obviously even (except for $a=0$, but then $b=0$, and $\frac{0}{0}$ is an indeterminate answer, and not rational). As an odd number cannot be equal to an even number, $2^{a}\neq 7^{b}$, and thus $\text{log}_{2}(7)\neq\frac{a}{b}$ and is not rational. This can be done for any number that is different in sign then the base of the logarithm, or by using modular arithmetic, any number with prime factors not in the base of the logarithm.