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

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We have to solve : log(4) (x+4) + log(4) 1/(x+1) - 1 = 0

Use the relation log a + log b = log (a*b)

log(4) (x+4) + log(4) 1/(x+1) - 1 = 0

=> log(4) (x + 4) / (x + 1) - 1 = 0

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

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

=> x + 4 = 4(x + 1)

=> x + 4 = 4x + 4

=> 3x = 0

=> x = 0

**Log 4 and log 1 exist, so x = 0 is a valid solution.**

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.**