# Determining the domain of a logarithmic equation

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.

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.

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.