Verify if the logarithmic equation has real solutions ? ln(x^2-1)-ln(x-1)=ln(4)
The quadratic equation ax^2 + bx + c = 0 has real solutions if b^2 >= 4ac
ln(x^2-1) - ln(x-1) = ln(4)
use the property ln a - ln b = ln (a/b)
=> ln[(x^2 - 1)/(x - 1)] = ln 4 is now converted to the quadratic equation:
=> (x^2 - 1)/(x - 1) = 4
=> x^2 - 1 = 4x - 4
=> x^2 - 4x + 3 = 0
b^2 = 4^2 = 16
4*a*c = 4*1*3 = 12
As 16 > 12 , b^2 > 4ac and the quadratic equation x^2 - 4x + 3 = 0 has real roots.
This implies that ln(x^2-1) - ln(x-1) = ln(4) has real roots.
We'll start by imposing constraints for the existence of the logarithmic functions:
x^2 - 1 > 0
x - 1 > 0
x > 1
The common interval of values of x that make the logarithmic functions to exist is (1 , +infinite).
We'll solve the equation applying the quotient property to the left side:
ln (x^2 - 1)/(x-1) = ln 4
We'll re-write the difference of 2 squares from numerator:
x^2 - 1 = (x-1)(x+1)
ln (x-1)(x+1)/(x-1) = ln 4
We'll simplify and we'll get:
ln (x + 1) = ln 4
Since the bases are matching, we'll apply one to one rule:
x + 1 = 4
x = 4 - 1
x = 3
Since the value of x belongs to the range of admissible values, we'll accept x = 3 as a solution of the equation.