Determining the domain of a logarithmic equation

Question

I constructed two examples of determining the doamin of a logarithmic equation:

Example 1

$$\ln\left(3\,a-x\right)\,\ln\left(2\,x+2\,a-5\right)=\ln\left(3\,a-x\right)\,\ln\left(x-a\right)\\ \ln\left(3\,a-x\right)\,\left(\ln\left(2\,x+2\,a-5\right)-\ln\left(x-a\right)\right)=0\\ \ln\left(3\,a-x\right)=0\quad \textbf{or} \quad\ln\left(2\,x+2\,a-5\right)-\ln\left(x-a\right)=0$$

For the right-side equation $$\ln\left(2\,x+2\,a-5\right)=\ln\left(x-a\right),$$ we should insist that $2x+2a-5>0\;$ AND $\;x-a>0.$

Example 2

$$\log_{3}\left(a-x^{2}\right)=\log_{3}\left(a-y^{2}\right)\\ a-x^{2}=a-y^{2}$$

Here, it should be enough to insist that $a-x^{2}>0\;$ OR $\;a-y^{2}>0$, because these both sides are equal, and if one of them is greater than zero then the other one is automatically also greater than zero.

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Why, with the first example, must we use AND, yet with second example, it is enough to use OR? I tried using the OR method in the first example, but it didn't work.

Answer 1

Example 1 $$\ln\left(2\,x+2\,a-5\right)=\ln\left(x-a\right),$$ we should insist that $2x+2a-5>0\;$ AND $\;x-a>0.$

Example 2 $$a-x^{2}=a-y^{2}$$ Here, it should be enough to insist that $a-x^{2}>0\;$ OR $\;a-y^{2}>0$

In fact, the above 'AND' and 'OR' can both be dropped (just choose the simpler expression to specify as positive), because generally, \begin{align}&\color{red}{\ln f(x)=\ln g(x)} \\\iff{}& f(x)=g(x)\quad \text{and}\quad f(x)>0\quad \text{and}\quad g(x)>0 \\\color{red}{\iff}{}&\color{red}{f(x)=g(x)\quad \text{and}\quad f(x)>0.}\end{align}

Using Example 1 to illustrate this:

• \begin{align}&\ln\left(2x+2a-5\right)=\ln\left(x-a\right)\\ \iff {}&2x+2a-5=x-a\quad\text{and}\color{violet}{\quad x-a>0}\\ \iff {}&x=5-3a\quad\text{and}\quad x>a\\ \iff {}&x=5-3a\quad\text{and}\quad 5-3a>a\\ \iff {}&x=5-3a\quad\text{and}\quad a<1.25\end{align}
• \begin{align}&\ln\left(2x+2a-5\right)=\ln\left(x-a\right)\\ \iff {}&2x+2a-5=x-a\quad\text{and}\quad \color{violet}{2x+2a-5>0}\\ \iff {}&x=5-3a\quad\text{and}\quad x>2.5-a\\ \iff {}&x=5-3a\quad\text{and}\quad 5-3a>2.5-a\\ \iff {}&x=5-3a\quad\text{and}\quad a<1.25\end{align}
• Desmos demonstration

Of possible interest: the bottom example at this Answer.