What is x if lg(x+1) = lg10 + 2lg(x-1) ?

First, we'll impose the constraints of existence of
logarithms:

x+1>0

x>-1

x-1>0

x>1

The
interval of admissible values for x is (1,+inf).

Now, we'll
solve the equation:

lg(x+1) = lg10 +
2lg(x-1)

We'll apply the power rule of logarithms for the
term 2lg(x-1):

2lg(x-1) = lg
(x-1)^2

We'll re-write the
equation:

lg(x+1) = lg10 +
lg(x-1)^2

Because the bases of the logarithms from
the right side are matching, we'll apply the product rule of
logarithms: 

lg a + lg b = lg
a*b

We'll put a = 10 and b =
(x-1)^2

lg(x+1) =
lg10(x-1)^2

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

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

We'll expand the square from the right
side:

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

We'll remove the
brackets:

x+1 = 10x^2 - 20x +
10

We'll subtract x+1 both sides and we'll apply symmetric
property:

10x^2 - 20x + 10 - x - 1 =
0

10x^2 - 21x + 9 = 0

We'll
apply the quadratic formula:

x1 =
[21+sqrt(441-360)]/20

x1 =
(21+9)/20

x1 = 30/20

x1 = 3/2
=1.5 > 1

x2 =
(21-9)/20

x2 = 12/20

x2 = 3/5
= 0.6 < 1

Since the second root
doesn't belong to the interval of admissible value, this one will be rejected and the
equation will have only one root, x = 1.5.