1. Solve $5^{-x} \cdot 5^{x-2} = \frac{25^{2x}}{5}$ Combine powers on the left: $5^{-x + x - 2} = 5^{-2}$ Rewrite right side: $25^{2x} = (5^2)^{2x} = 5^{4x}$, so right side is $\frac{5^{4x}}{5} = 5^{4x - 1}$ Set exponents equal: $-2 = 4x - 1$ Solve: $4x = -1$, $x = -\frac{1}{4}$ 2. Solve $2^{x+4} + 2^x = 8704$ Rewrite $2^{x+4} = 2^4 \cdot 2^x = 16 \cdot 2^x$ Let $y = 2^x$, then $16y + y = 17y = 8704$ Solve: $y = \frac{8704}{17} = 512$ Since $2^x = 512 = 2^9$, $x = 9$ 3. Solve $9^x + 9 = 10 \cdot 3^x$ Rewrite $9^x = (3^2)^x = 3^{2x}$ Let $y = 3^x$, then $y^2 + 9 = 10y$ Rewrite as quadratic: $y^2 - 10y + 9 = 0$ Factor: $(y - 9)(y - 1) = 0$ Solutions: $y = 9$ or $y = 1$ Back-substitute: $3^x = 9 = 3^2 \Rightarrow x=2$ or $3^x=1=3^0 \Rightarrow x=0$ 4. Solve $\sqrt{7x - 12} - x = 0$ Rewrite: $\sqrt{7x - 12} = x$ Square both sides: $7x - 12 = x^2$ Rearranged: $x^2 - 7x + 12 = 0$ Factor: $(x - 3)(x - 4) = 0$ Solutions: $x=3$ or $x=4$ Check in original equation: For $x=3$: $\sqrt{21 - 12} - 3 = \sqrt{9} - 3 = 3 - 3 = 0$ valid For $x=4$: $\sqrt{28 - 12} - 4 = \sqrt{16} - 4 = 4 - 4 = 0$ valid 5. Solve $(\frac{1}{6})^{3x+2} \cdot 216^{3x} = \frac{1}{216}$ Rewrite bases: $\frac{1}{6} = 6^{-1}$, $216 = 6^3$ Expression becomes: $6^{-1(3x+2)} \cdot (6^3)^{3x} = 6^{-3}$ Simplify exponents: $6^{-3x - 2} \cdot 6^{9x} = 6^{-3}$ Combine: $6^{(-3x - 2 + 9x)} = 6^{-3}$ Simplify exponent: $6^{6x - 2} = 6^{-3}$ Set exponents equal: $6x - 2 = -3$ Solve: $6x = -1$, $x = -\frac{1}{6}$ 6. Solve $x + 3\sqrt{x+2} = 2$ Let $y = \sqrt{x+2}$, then $x = y^2 - 2$ Substitute: $y^2 - 2 + 3y = 2$ Rearranged: $y^2 + 3y - 4 = 0$ Factor: $(y + 4)(y - 1) = 0$ Solutions: $y = -4$ (discard, since $y=\sqrt{\cdot} \geq 0$), $y=1$ Back-substitute: $\sqrt{x+2} = 1 \Rightarrow x+2=1 \Rightarrow x=-1$ 7. Solve $\sqrt{2} - x - x = -2$ Rewrite: $\sqrt{2} - 2x = -2$ Isolate $x$: $-2x = -2 - \sqrt{2}$ Divide: $x = \frac{2 + \sqrt{2}}{2} = 1 + \frac{\sqrt{2}}{2}$ 8. Solve $2^{2x} + 2^x - 6 = 0$ Let $y = 2^x$, then $y^2 + y - 6 = 0$ Factor: $(y + 3)(y - 2) = 0$ Solutions: $y = -3$ (discard, $2^x > 0$), $y=2$ Back-substitute: $2^x = 2 = 2^1 \Rightarrow x=1$ 9. Solve $2^x + 2^{x+2} = 40$ Rewrite $2^{x+2} = 2^2 \cdot 2^x = 4 \cdot 2^x$ Let $y = 2^x$, then $y + 4y = 5y = 40$ Solve: $y = 8$ Back-substitute: $2^x = 8 = 2^3 \Rightarrow x=3$