What are the potential solutions of ln(x^2 -25)= 0

\(\log_ab=c\iff a^c=b,\ \text{for}\ a>0,\ a eq1,\ b>0\\\\\text{Domain:}\\x^2-25>0\\x^2-5^2>0\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\(x-5)(x+5)>0\to x<-5\ \vee\ x>5\\D:x\in(-\infty,\ -5)\ \cup\ (5,\ \infty)\\\\\ln(x^2-25)=0\iff x^2-25=e^0\\\\\text{we know}\ a^0=1\ \text{for all values of}\ a\ \text{except 0}\\\\x^2-25=1\qquad\text{add 25 to both sides}\\x^2=26\to x=\pm\sqrt{26}\\\\x=-\sqrt{26}\in D\\\\x=\sqrt{26}\in D\)

eudora eudora · Answer 2 · 2019-10-31T23:07:01-04:00

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

\(ln_{e}(x^{2}-25)=0\) [Since base of a natural logarithm is considered as e]

\(e^{0}=(x^{2}-25)\)

Since in a logarithmic function if \(ln_{e}b=c\)

Then \(e^{c}=b\)

Now we further solve the expression to get the possible roots.

\(x^{2}-25=1\)

\(x^{2}=25+1\)

\(x^{2}=26\)

x = ±√26

Therefore, the potential solutions are x = √26 and x = -√26.