Answer:
there are no real solutions
Step-by-step explanation:
[tex]\log_5(x-1)+\log_5(x+3)-1=0\\\\\log_5(x-1)+\log_5(x+3)=1[/tex]
there is a rule that says
[tex]\log_b(a)+\log_b(c)=\log_b(ac)[/tex]
so we have
[tex]\log_5(x-1)+\log_5(x+3)=\log_5((x-1)(x+3))=1[/tex]
and we have the definition
[tex]\log_a(b)=c\\\\a^{c} =b[/tex]
so we have
[tex]5^{1} = (x-1)(x+3)\\\\5=x^{2} -1x+3x-3\\\\5=x^{2} +2x-3\\\\0=x^{2} +2x-3-5\\\\0=x^{2} +2x-8\\\\[/tex]
and using the quadratic formula we get that
there are no real solutions
