Verify if the logarithmic equation has real solutions ? ln(x^2-1)-ln(x-1)=ln(4)

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justaguide | College Teacher | (Level 2) Distinguished Educator

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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.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

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