Question

Log5(x-1)+log5(x+3)-1=0

Answer 1

Answer:

There is no answer for this

Step-by-step explanation:

Answer 2

Answer:

there are no real solutions

Step-by-step explanation:

$\log_5(x-1)+\log_5(x+3)-1=0\\\\\log_5(x-1)+\log_5(x+3)=1$

there is a rule that says

$\log_b(a)+\log_b(c)=\log_b(ac)$

so we have

$\log_5(x-1)+\log_5(x+3)=\log_5((x-1)(x+3))=1$

and we have the definition

$\log_a(b)=c\\\\a^{c} =b$

so we have

$5^{1} = (x-1)(x+3)\\\\5=x^{2} -1x+3x-3\\\\5=x^{2} +2x-3\\\\0=x^{2} +2x-3-5\\\\0=x^{2} +2x-8$

and using the quadratic formula we get that

there are no real solutions