Log 5 (5x+25) - 2 = 0

For starting, we'll have to impose constraints of
existance of logarithm
function.

5x+25>0

We'll
add -25 both
sides:

5x>-25

We'll
divide by
5:

x>-5

So, for the
logarithms to exist, the values of x have to be in the interval (-5,
+inf.)

We'll shift the free term to the right
side:

log 5 (5x+25) = 2

We'll
create matching bases to the right side.

log 5 (5x+25) =
log 5 (5^2)

Now, we'll use the one to one
property:

5x+25 = 25

We'll
eliminate like terms:

5x =
25-25

5x = 0

We'll divide by
5:

x = 0 >
-5

The solution is admissible because the
value belongs to the interval (-5,+inf.)