Answer :
Answer:
[tex]\large\boxed{x=-\sqrt{26}\ \vee\ x=\sqrt{26}}[/tex]
Step-by-step explanation:
[tex]\log_ab=c\iff a^c=b,\ \text{for}\ a>0,\ a\neq1,\ b>0\\\\\text{Domain:}\\x^2-25>0\\x^2-5^2>0\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\(x-5)(x+5)>0\to x<-5\ \vee\ x>5\\D:x\in(-\infty,\ -5)\ \cup\ (5,\ \infty)\\\\\ln(x^2-25)=0\iff x^2-25=e^0\\\\\text{we know}\ a^0=1\ \text{for all values of}\ a\ \text{except 0}\\\\x^2-25=1\qquad\text{add 25 to both sides}\\x^2=26\to x=\pm\sqrt{26}\\\\x=-\sqrt{26}\in D\\\\x=\sqrt{26}\in D[/tex]
Answer:
The potential solutions are x = √26 and x = -√26
Step-by-step explanation:
The given expression is ln(x² - 25) = 0
It can be rewritten as
[tex]ln_{e}(x^{2}-25)=0[/tex] [Since base of a natural logarithm is considered as e]
[tex]e^{0}=(x^{2}-25)[/tex]
Since in a logarithmic function if [tex]ln_{e}b=c[/tex]
Then [tex]e^{c}=b[/tex]
Now we further solve the expression to get the possible roots.
[tex]x^{2}-25=1[/tex]
[tex]x^{2}=25+1[/tex]
[tex]x^{2}=26[/tex]
x = ±√26
Therefore, the potential solutions are x = √26 and x = -√26.