# Logarithms.Find x for lg(x)+lg(x+1)=(lg90) to exist.

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### 2 Answers

We need to find x for lg(x)+lg(x+1)=(lg90) to exist.

Lets simplify the expression:

lg(x) + lg(x+1) = lg(x*(x + 1)) = (lg 90)

x(x + 1) = 90

x^2 + x = 90

=> x^2 + x - 90 = 0

=> x^2 + 10x - 9x - 90 = 0

=> x(x + 10) - 9(x + 10) = 0

=> (x - 9)(x + 10) = 0

x = 9 and x = -10

As the log of negative numbers does not exist we eliminate x = -10.

**The required solution of the equation is x = 9**

We'll impose the constraints of existence of logarithms:

x > 0

x + 1 > 0

x > -1

The range of admissible values for x is (0 ; +infinite).

Since the bases are matching, we'll transform the sum from the left side into a product:

lg x(x+1) = lg 90

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

x(x+1) = 90

We'll remove the brackets:

x^2 + x - 90 = 0

We'll apply the quadratic formula:

x1 = [-1+sqrt(1+360)]/2

x1 = (-1+19)/2

x1 = 9

x2 = (-1-19)/2

x2 = -10 (we'll reject it)

**The only valid solution of the equation is x = 9.**