Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 342: 21

$x=-1$ or $x=3$

$2^{x^2-2x}=8$ Expressing the right side in terms of an exponent with base 2: $2^{x^2-2x}=2^3$ Taking base-2 log of both sides: $x^2-2x=3$ Thus, $x^2-2x-3=0$ Factorising the right side: $(x+1)(x-3)=0$ Therefore, $x=-1$ or $x=3$.