Answer:
(a) x = 12 only
Step-by-step explanation:
You want the solution to 1/2ln(x+37) = ln(x^2−25)−ln(x+5).
Graph
A graph of this written in the form ...
[tex]\dfrac{1}{2}\ln{(x+37)}-(\ln{(x^2-25)}-\ln{(x+5)})=0[/tex]
shows one solution at x=12.
Antilog
The right side of the equation can be simplified to ...
[tex]\dfrac{1}{2}\ln(x+37)=\ln{\left(\dfrac{x^2-25}{x+5}\right)}=\ln{(x-5)}\\\\x+37=(x-5)^2\qquad\text{multiply by 2; take antilogs}\\\\x^2-11x-12=0\qquad\text{subtract $x+37$}\\\\(x+1)(x-12)=0\qquad\text{factor}[/tex]
The argument of the ln function must be positive, so we require x > 5. The factors will be zero for x=-1 and x=12, but x=-1 is not in the domain of the original equation.
The only solution is x = 12.
