**Answer:**

[tex]x=\sqrt{26}[/tex] and [tex]x=-\sqrt{26}[/tex].

**Step-by-step explanation:**

The given equation is

[tex]\ln(x^2-25)=0[/tex]

We need to find he potential solution of the given equation.

Potential solution of an equation are the solutions of that equation which are lie on its domain. So, first find the domain of given equation.

Since logarithmic function is defined for positive values, therefore

[tex]x^2-25>0[/tex]

[tex]x^2>25[/tex]

It means x<-5 or x>5. So domain of the equation is

Domain = [tex](-\infty,-5)\cup (5,\infty)[/tex]

We know that ln(1) = 0. The given equation can be written as

[tex]\ln(x^2-25)=\ln 1[/tex]

On comparing both sides we get

[tex]x^2-25=1[/tex]

Add 25 on both sides.

[tex]x^2=26[/tex]

Taking square root on both sides.

[tex]x=\pm \sqrt{26}[/tex]

Both [tex]\sqrt{26}[/tex] and [tex]-\sqrt{26}[/tex] belongs to the domain.

Therefore, the potential solutions of the given equation are [tex]x=\sqrt{26}[/tex] and [tex]x=-\sqrt{26}[/tex].