Answer:
[tex] x =\pm \sqrt{26}[/tex]
Step-by-step explanation:
Given :-
And we need to find the potential solutions of it. The given equation is the logarithm of x² - 25 to the base e . e is Euler's Number here. So it can be written as ,
Equation :-
[tex]\implies log_e {(x^2-25)}= 0 [/tex]
In general :-
- If we have a logarithmic equation as ,
[tex]\implies log_a b = c [/tex]
Then this can be written as ,
[tex]\implies a^c = b [/tex]
In a similar way we can write the given equation as ,
[tex]\implies e^0 = x^2 - 25 [/tex]
- Now also we know that [tex]a^0 = 1[/tex] Therefore , the equation becomes ,
[tex]\implies 1 = x^2 - 25 \\\\\implies x^2 = 25 + 1 \\\\\implies x^2 = 26 \\\\\implies x =\sqrt{26} \\\\\implies x = \pm \sqrt{ 26}[/tex]
Hence the Solution of the given equation is ±√26.
