If $\frac{log 2}{log 3}$ is irrational how does it follow that both $1-\frac{1}{2^s}$ and $1-\frac{1}{3^s} $ can't be zero for the same s

Question

Wikipedia page https://en.m.wikipedia.org/wiki/Dirichlet_eta_function says:

[N]ow we can define correctly, where the denominators are not zero,

$\zeta(s) = \frac{\eta(s)}{1-\frac{2}{2^s}}$

or

$\zeta(s) = \frac{\lambda(s)}{1-\frac{3}{3^s}}$

Since $ \frac{\log 3}{\log 2} $ is irrational, the denominators in the two definitions are not zero at the same time except when s=1 ...

How does it follow? Any pointers to more detail appreciated.

Follow up question: I can see that both the denominators cannot be zero for the same value $s \in \Bbb C (s \ne 1)$, but can any value of $s$ with these constraints make any one of the above denominators zero at all?

Considering $$1-3^{(1-s)} = 0$$ $$\Rightarrow 3^{(1-s)} = 1$$ $$\Rightarrow (1-s) \log 3 = 0$$ $$\Rightarrow (1-s) = 0$$ $$\Rightarrow 1 = s$$ which is excluded from the domain.

In other words, does $ 3^{1-s} = 1 $ have any other roots other than $s = 1$ when $s \in \Bbb C$

Answer 1

If the denominators are zero, then it follows that: $$1-2^{1-s}=1-3^{1-s} $$ $$\implies \left(\frac 23 \right) ^{1-s} =1$$

After taking $\log$ on both sides: $$\implies \,? $$

Note that we have: $$\eta(s) = \frac{1-2^{1-s}}{1-3^{1-s}} \lambda(s) $$ which means that if $s \neq 1$, we get that $$\eta(s) =0$$