1. **Problem 1:** Find $x$ such that $2^{2x-3} = 8$. 2. We know that $8$ can be written as a power of $2$: $8 = 2^3$. 3. So the equation becomes: $$2^{2x-3} = 2^3$$ 4. Since the bases are the same and non-zero, we can set the exponents equal: $$2x - 3 = 3$$ 5. Solve for $x$: $$2x = 3 + 3$$ $$2x = 6$$ $$x = \frac{6}{2}$$ $$x = 3$$ --- 1. **Problem 2:** Find $x$ such that $3^{3x} \times 3^{x} = 81$. 2. Use the rule of exponents: $a^m \times a^n = a^{m+n}$. 3. So, $$3^{3x} \times 3^{x} = 3^{3x + x} = 3^{4x}$$ 4. Rewrite $81$ as a power of $3$: $$81 = 3^4$$ 5. Set the exponents equal: $$4x = 4$$ 6. Solve for $x$: $$x = \frac{4}{4}$$ $$x = 1$$