1. State the problem: Solve the inequality $$2^{x-1} \geq 8$$. 2. Express 8 as a power of 2: Since $$8 = 2^3$$, rewrite the inequality as $$2^{x-1} \geq 2^3$$. 3. Since the base 2 is greater than 1, the inequality direction remains the same when taking logarithms or comparing exponents. So, we have: $$x - 1 \geq 3$$. 4. Solve for $$x$$: $$x \geq 3 + 1$$ $$x \geq 4$$. Final answer: $$x \geq 4$$.