Question:

Step 1: Introduce a substitution to simplify the equation. Let $t = 2^x$. This substitution will help us rewrite the exponential expressions in terms of $t$. Using this substitution: $$4^x = (2^2)^x = (2^x)^2 = t^2$$ $$2^{x+1} = 2 \cdot 2^x = 2t$$ $$2^{x+3} = 2^3 \cdot 2^x = 8t$$ Step 2: Rewrite the original equation in terms of $t$. Substituting these expressions into the original equation: $$\sqrt{t^2 - 2t + 1} + \sqrt{t^2 - 8t + 16} = 3$$ Step 3: Recognize and simplify perfect square trinomials. Notice that both expressions under the square roots are perfect squares: $$t^2 - 2t + 1 = (t-1)^2$$ $$t^2 - 8t + 16 = (t-4)^2$$ Therefore, the equation becomes: $$\sqrt{(t-1)^2} + \sqrt{(t-4)^2} = 3$$ Step 4: Apply the definition of absolute value. Since $\sqrt{a^2} = |a|$ for any real number $a$: $$|t-1| + |t-4| = 3$$ Step 5: Analyze the absolute value equation using the triangle inequality. By the triangle inequality property, for any real numbers $a$ and $b$: $$|a - b| \leq |a| + |b|$$ In particular, we have: $$|t-1| + |t-4| \geq |(t-1) - (t-4)| = |3| = 3$$ Equality holds in the triangle inequality when the terms have the same sign, which occurs when: $$1 \leq t \leq 4$$ Step 6: Determine the solution set for $t$. Since $|t-1| + |t-4| = 3$ and the minimum value of $|t-1| + |t-4|$ is exactly $3$ (achieved when $1 \leq t \leq 4$), the equation is satisfied for all: $$t \in [1, 4]$$ Step 7: Convert back to the original variable $x$. Since $t = 2^x$, we need: $$2^x \in [1, 4]$$ Taking logarithm base 2: $$x \in [\log_2 1, \log_2 4]$$ $$x \in [0, 2]$$ **Final Answer:** The solution set is $x \in [0, 2]$, which corresponds to **Option 3**.