Quadratic exponent

$3^{x^2 -2x -5} = 27 =3^3$

$\implies x^2 -2x -5 =3$

$\implies x=\{-2,4\}$

$\implies x_1 =4 , |x_2| = 2$

$\dfrac{\ln |x_2|}{\ln x_1} = \dfrac{\ln 2}{\ln 4} =\log_{4} 2 = \boxed{0.5}$

We know that $27 = 3^3$ : $3^{x^2-2x-5}=27$ $\Longrightarrow3^{\blue{x^2-2x-5}}=3^{\blue{3}}$ Since the bases are the same, we can deduce that the exponents must be the same: $\Longrightarrow x^2-2x-5 = 3$ $\xRightarrow[]{\text{subtract }3} x^2-2x-8=0$ $\xRightarrow[]{\text{factorise}} (x+2)(x-4)=0$ We can see that our solutions for $x$ are $4$ and $-2$ and since $4 > -2$ we can say that $\boxed{x_1 = 4}$ and $\boxed{x_2 = -2}$ . Now, we can work out the answer: $\large \frac{\ln|x_2|}{\ln x_1} = \frac{\ln|-2|}{\ln 4} = \frac{\ln 2}{2\ln 2} = \green{\boxed{\frac{1}{2}}}$