# ln(-x)=ln(x^2-6)How to find out the value of x in the above problem..please help

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### 2 Answers

The equation to be solved for x is : ln(-x) = ln(x^2 - 6)

ln(-x) = ln(x^2 - 6)

=> -x = x^2 - 6

=> x^2 + x - 6 = 0

=> x^2 + 3x - 2x - 6 = 0

=> x(x + 3) - 2(x + 3) = 0

=> (x - 2)(x + 3) = 0

x = 2 and x = -3

But ln -2 is not defined. So the only solution is x = -3.

**The value of x that satisfies the equation is x = -3**

Since the bases are matching, we'll apply one to one rule of logarithms:

-x = x^2 - 6

We'll use the symmetrical property and we'll have:

x^2 - 6 = -x

We'll shift -x to the left:

x^2 + x - 6 = 0

We'll apply quadratic formula:

x1 = [-1+sqrt(1+24)]/2

x1 = (-1+5)/2

x1 = -2

x2 = -3

Now, we'll impose the constraints of existence of logarithms:

x < 0

x^2 - 6 > 0

The expression is positive if x belongs to the intervals: (-infinite;-sqrt6)U(+sqrt6 ; +infinite)

The common interval of admissible values for x is (-infinite;-sqrt6).

**Since x = -2 doesn't belong to this interval, the only solution of the equation is x = -3.**