Determining Irrationality Using Logarithms?

Using the properties of logarithms have that

$\log_{4}(18) = \log_{4}(2*3^{2}) = \log_{4}(2) + 2\log_{4}(3) = \dfrac{1}{2} + 2*\dfrac{\log_{2}(3)}{\log_{2}(4)} = \dfrac{1}{2} + \log_{2}(3)$ .

Now this expression will be rational if $\log_{2}(3) \gt 0$ is rational. So by contradiction, assume that $\log_{2}(3) = \dfrac{m}{n}$ for some positive integers $m,n$ . This then implies that

$\large 2^{\frac{m}{n}} = 3 \Longrightarrow (2^{\frac{m}{n}})^{n} = 3^{n} \Longrightarrow 2^{m} = 3^{n}$ .

But as $2^{m}$ is always even and $3^{n}$ is always odd, this equality can never hold. This implies that our supposition that $\log_{2}(3)$ is rational is in fact false, i.e., $\log_{2}(3)$ is irrational. Then since the sum of a rational and an irrational is irrational the correct option is $\boxed{irrational}$ .

The next question to consider is whether $\log_{2}(3)$ , and thus $\log_{4}(18)$ , is algebraic or transcendental. By the Gelfond-Schneider Theorem , if the irrational number $\log_{2}(3)$ were algebraic then $\large 2^{\log_{2}(3)} = 3$ would be transcendental, which is clearly not the case. Thus $\log_{2}(3)$ must be transcendental.

Provided that $16 < 18 < 64$ , we get that $2 < \log_4 18 < 3$ . This is obvious that it cannot be prime, composite or rational, hence it is irrational

Let log4(18) = x ==> 4^x = 18 ==> 2^(2x) = 18 ==> x is irrational.

first of all, here is a little exam trick I learned, we know only one answer is true

if the answer was prime then it's rational making two answers correct therefore it can't be prime , similarly, it can't be composite .

now we need to prove whether it's rational or not, we do a proof by contradiction using a method similar to what other solvers had done, assuming it's rational and coming up with a contradiction proving it's irrational .