# LogarithmsFind x if lg(x+1) = lg10+2lg(x-1).

giorgiana1976 | Student

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

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.