How to find the real x in equation log4 (x+4)+log4 1/(x+1) -1=0?

First, we'll impose conditions of existence of
the logarithms.

x+4>0

x>-4

and

x+1>0

x>-1

The
range of admissible values, for the logarithms to exist is: (-1,
+inf).

The equation is a sum of 2 logarithms with the same
base, 4, and, according to the rule, the sum of 2 logarithms with the same base is
transforming in the logarithm of the product.

log4
(x+4)+log4 1/(x+1) = log4 (x+4)*[1/(x+1)]

We'll move 1 to
the right side:

log4 (x+4)*[1/(x+1)] =
1

log4 (x+4)/(x+1) =1

We'll
take antilogarithm:

(x+4)/(x+1) =
4^1

(x+4)/(x+1) =
4

(x+4)=4*(x+1)

We'll remove
the
brackets:

x+4=4x+4

x-4x+4-4=0

-3x=0

x=0

Since
x = 0 belongs to the set (-1, +inf), then x=0.