# Solve for real x:log(5) (x+4) = log(5) (5x+5)

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log(5) (x+4) = log(5) (5x+5).

We are required to solve for x:

Since both sides have the same base of the logarithms, we take antilog with respect to the base 5.

x+4 = 5x+5.

We isolate x to left and numbers to right.

x-5x = 5-4

-4x = 1

x = 1/-4.

x = 1/4.

**So x= -1/4.**

Before solving the equation, we'll impose conditions of existence of the logarithms.

x+4>0

x>-4

and

5(x+1)>0

x>-1

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

We notice that the logarithms have matching bases, so we can apply the one to one property:

x+4=5x+5

We'll move all terms to one side:

x-5x+4-5=0

-4x - 1=0

We'll add 1 both sides:

-4x = 1

x = -1/4

x=-0.25

After finding the value for x, we'll have to check if it is a solution for the equation, so, we'll have to verify if it is belonging to the range of values (-1, +inf). We notice that -0.25 is belonging to the interval (-1, +inf).

**x=-0.25**