Fake log formula

We can write $\log(A)+\log(B)+\log(C)=\log(A\cdot B\cdot C)$ .

So, the equation is

$\begin{aligned} &&\log(A+B+C)&=\log(A)+\log(B)+\log(C)\\ &\Leftrightarrow&\log(A+B+C)&=\log(A\cdot B\cdot C)\\ &\Leftrightarrow&A+B+C&=A\cdot B\cdot C \end{aligned}$

So, we need to find three distinct positive integers where the product is equal to the sum.

As we don't count permutations, let's say $1\leq A<B<C$ and we write the sorted triplet as $(A,B,C)$ .

The smallest triplet is $(1,2,3)$ :

$1+2+3=6=1\cdot 2\cdot 3$ ,

so we already found one valid triplet.

If at least one of the numbers gets greater by $n\geq 1$ , the sum gets greater by $n$ , too, while the product is increased by at least $2n$ , as $1<B<C$ .

So, $(1,2,3)$ is the only triplet that satisfies the equation $\log(A+B+C)=\log(A)+\log(B)+\log(C)$