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