log_4(-x)+log_4(x+10)=2 Solve the equation. Check for extraneous solutions.

To evaluate the given equation log_4(-x)+log_4(x+10)=2 , we may apply the logarithm property: log_b(x)+log_b(y)=log_b(x*y) .
log_4(-x)+log_4(x+10)=2
log_4((-x)*(x+10))=2
log_4(-x^2-10x)=2
To get rid of the "log" function, we may apply the logarithm property: b^(log_b(x))=x.
Raise both sides by base of 4 .
4^(log_4(-x^2-10x))=4^2
-x^2-10x=16
Add x^2 and 10x on both sides of the equation to simplify in standard form: ax^2+bx+c= 0.
-x^2-10x+x^2+10x=16+x^2+10x
0=16+x^2+10x orx^2+10x+16=0.
Apply factoring on the trinomial.
(x+2)(x+8)=0
Apply zero-factor property to solve for x by equating each factor to 0 .
x+2=0
x+2-2=0-2
x=-2
and
x+8=0
 x+8-8=0-8 
x=-8
Checking: Plug-in each x on log_4(-x)+log_4(x+10)=2 .
Let x=-2 on  log_4(-x)+log_4(x+10)=2 .
log_4(-(-2))+log_4(-2+10)=?2
log_4(2)+log_4(8)=?2
log_4(2*8)=?2
log_4(16)=?2
log_4(4^2)=?2
2log_4(4)=?2
2*1=?2
2=2        TRUE
Let x=-8 on log_4(-x)+log_4(x+10)=2.
log_4(-(-8))+log_4(-8+10)=?2
log_4(8)+log_4(2)=?2
log_4(8*2)=?2
log_4(16)=?2
2=2        TRUE
Therefore, there are no extraneous solutions.
Both solved x-values: x=-2 and x=-8 are real solution of the equation log_4(-x)+log_4(x+10)=2 .