Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 352: 16

$x=405.429$ $x=-401.429$

$ln((x-2)^2)=12$ $e^{ln((x-2)^2)}=e^{12}$ $(x-2)^2=e^{12}$ $x^2-4x+4-e^{12}=0$ //use ABC formula $x=\frac{4\pm\sqrt{16-4(1)(4-e^{12})}}{2}$ $=2\pm\sqrt{e^{12}}$ $x=2+\sqrt{e^{12}}$ $x \approx 405.428793$ $=405.429$ $x=2-\sqrt{e^{12}}$ $x \approx -401.428793$ $=-401.429$