Find the number of elements in the set $$\{(a, b) \in\mathbb{N}: 2\le a, b < 2023, \log_{a}(b)+6\log_{b}(a)=5\}$$.

The condition given is $$\log_{a}(b)+6\log_{b}(a)=5$$ with $$2\le a,b\lt 2023$$ and $$a,b\in\mathbb{N}$$.

Introduce the variable $$x=\log_{a}(b)$$. By the reciprocal property of logarithms, $$\log_{b}(a)=\frac{1}{x}$$ because $$\log_{b}(a)=\frac{1}{\log_{a}(b)}$$.

Substituting these into the equation gives
$$x+6\left(\frac{1}{x}\right)=5$$

Multiply by $$x$$ to clear the denominator:
$$x^{2}-5x+6=0$$

Factorisation gives
$$(x-2)(x-3)=0$$
so $$x=2$$ or $$x=3$$.

$$x=2\; \Rightarrow\;\log_{a}(b)=2\;\Rightarrow\;b=a^{2}$$
Constraints: $$2\le a\lt 2023$$ and $$b=a^{2}\lt 2023$$.
We need $$a^{2}\lt 2023$$, so $$a\lt\sqrt{2023}\approx 44.97$$.
Thus $$a$$ can take integer values $$2,3,\dots,44$$ — a total of $$44-2+1=43$$ choices, and each such $$a$$ gives exactly one corresponding $$b=a^{2}$$.

$$x=3\; \Rightarrow\;\log_{a}(b)=3\;\Rightarrow\;b=a^{3}$$
Constraints: $$b=a^{3}\lt 2023$$.
We need $$a^{3}\lt 2023$$, so $$a\lt\sqrt[3]{2023}\approx 12.63$$.
Thus $$a$$ can take integer values $$2,3,\dots,12$$ — a total of $$12-2+1=11$$ choices, each giving one $$b=a^{3}$$.

Adding both cases, the number of ordered pairs $$(a,b)$$ is $$43+11=54$$.

Hence the required number of elements in the set is 54.