3-lg(X^2+1)=lg(1/(x+1))+log5 5first 5 is little number, I guess that log5 5 equals 1.

Yes, it is true, log5 5 =
1.

We'llr e-write the given
equation:

3 - lg(x^2 + 1) = lg [1/(x+1)] +
1

We'll create matching bases and we'll start by writting 1
= lg 10.

3 - lg(x^2 + 1) = lg [1/(x+1)] + lg
10

We can write 3 = 3*1 = 3*lg
10

We'll use power
property:

3*lg 10 = lg
10^3

The equation will
become:

lg 10^3 - lg(x^2 + 1) = lg [1/(x+1)] + lg
10

We'll use the quotient property of logarithms to the
left side:

lg 1000/(x^2 + 1) = lg [1/(x+1)] + lg
10

We'll use the product property of logarithms to the
right side:

lg 1000/(x^2 + 1) = lg
[10/(x+1)]

Since the bases are matching, we'll apply one to
one property;

1000/(x^2 + 1) = 
10/(x+1)

We'll cross
multiply:

10(x^2 + 1) =
1000(x+1)

We'll divide by
10:

x^2 + 1 = 100x + 100

We'll
subtract 100x + 100 both sides:

x^2 - 100x - 99 =
0

x1 = [100+sqrt(100396)]/2

x1
= 50 + sqrt25099

x2 = 50 -
sqrt25099

Only the 1st value of x respects
the constraints of existence of logarithms, therefore the equation will have only one
solution x = 50 + sqrt25099.