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

Let $t={\log }_a(b)$. Since ${\log }_b(a)=1/t$, the equation becomes:

This gives $t=2$ or $t=3$, meaning:

We now count the number of valid pairs $(a,b)$ with $2\le a,b\le 2023$:

• Case 1: $b=a^2$. We require $2\le a\le 2023$ and $2\le a^2\le 2023⟹a\le \sqrt{2023}\approx 44.97$. Thus $a\in \{2,3,\dots ,44\}$, which gives $44-2+1=43$ valid pairs.

• Case 2: $b=a^3$. We require $2\le a\le 2023$ and $2\le a^3\le 2023⟹a\le \sqrt[3]{2023}\approx 12.65$. Thus $a\in \{2,3,\dots ,12\}$, which gives $12-2+1=11$ valid pairs.

Since $a^2\ne a^3$ for $a\ge 2$, there is no overlap between the two cases.
The total number of elements in the set is $43+11=54$.