(2) $$ 3 \cdot 5^{x}+5^{x+1}=40 $$ (4) $$ \sqrt{9^{x+2}}-3^{x-1}=\frac{26}{27} $$ (6) $$ 4^{x}+16=10 \cdot 2^{x} $$ (8) $$ 16^{x}-3 \cdot 4^{x+1}=-32 $$ (10) $$ 5^{x}-5^{2-x}=24 $$ (12) $$ 3^{4 x}-12 \cdot 3^{2 x}+27=0 $$

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Sure, let's solve each of the given equations step by step.

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### Problem (2)

**Equation:**
$$ 3 \cdot 5^{x} + 5^{x+1} = 40 $$

**Solution:**
1. Rewrite $$5^{x+1}$$ as $$5 \cdot 5^x$$:
   $$ 3 \cdot 5^{x} + 5 \cdot 5^{x} = 40 $$
2. Factor out $$5^x$$:
   $$ 5^{x} (3 + 5) = 40 $$
   $$ 5^{x} \cdot 8 = 40 $$
3. Solve for $$5^x$$:
   $$ 5^{x} = \frac{40}{8} = 5 $$
4. Since $$5^x = 5^1$$, we have:
   $$ x = 1 $$

**Answer:**
$$ x = 1 $$

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### Problem (4)

**Equation:**
$$ \sqrt{9^{x+2}} - 3^{x-1} = \frac{26}{27} $$

**Solution:**
1. Express $$9^{x+2}$$ in terms of base 3:
   $$ 9^{x+2} = (3^2)^{x+2} = 3^{2x + 4} $$
2. Take the square root:
   $$ \sqrt{3^{2x + 4}} = 3^{x + 2} $$
3. Rewrite the equation:
   $$ 3^{x + 2} - 3^{x - 1} = \frac{26}{27} $$
4. Factor out $$3^{x - 1}$$:
   $$ 3^{x-1} (3^3 - 1) = \frac{26}{27} $$
   $$ 3^{x-1} \cdot 26 = \frac{26}{27} $$
5. Solve for $$3^{x-1}$$:
   $$ 3^{x-1} = \frac{1}{27} = 3^{-3} $$
6. Equate the exponents:
   $$ x - 1 = -3 $$
   $$ x = -2 $$

**Answer:**
$$ x = -2 $$

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### Problem (6)

**Equation:**
$$ 4^{x} + 16 = 10 \cdot 2^{x} $$

**Solution:**
1. Express $$4^x$$ as $$2^{2x}$$:
   $$ 2^{2x} + 16 = 10 \cdot 2^{x} $$
2. Let $$y = 2^x$$, then $$2^{2x} = y^2$$:
   $$ y^2 + 16 = 10y $$
3. Rearrange the equation:
   $$ y^2 - 10y + 16 = 0 $$
4. Solve the quadratic equation:
   $$ y = \frac{10 \pm \sqrt{100 - 64}}{2} = \frac{10 \pm 6}{2} $$
   $$ y = 8 \quad \text{or} \quad y = 2 $$
5. Convert back to $$x$$:
   - For $$y = 8$$:
     $$ 2^x = 8 \Rightarrow x = 3 $$
   - For $$y = 2$$:
     $$ 2^x = 2 \Rightarrow x = 1 $$

**Answer:**
$$ x = 1 \quad \text{and} \quad x = 3 $$

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### Problem (8)

**Equation:**
$$ 16^{x} - 3 \cdot 4^{x+1} = -32 $$

**Solution:**
1. Express $$16^x$$ and $$4^{x+1}$$ in terms of base 4:
   $$ 16^{x} = (4^2)^x = 4^{2x} $$
   $$ 4^{x+1} = 4^x \cdot 4 = 4 \cdot 4^x $$
2. Let $$y = 4^x$$, then:
   $$ y^2 - 3 \cdot 4y = -32 $$
3. Rearrange the equation:
   $$ y^2 - 12y + 32 = 0 $$
4. Solve the quadratic equation:
   $$ y = \frac{12 \pm \sqrt{144 - 128}}{2} = \frac{12 \pm 4}{2} $$
   $$ y = 8 \quad \text{or} \quad y = 4 $$
5. Convert back to $$x$$:
   - For $$y = 8$$:
     $$ 4^x = 8 \Rightarrow (2^2)^x = 2^3 \Rightarrow 2^{2x} = 2^3 \Rightarrow 2x = 3 \Rightarrow x = \frac{3}{2} $$
   - For $$y = 4$$:
     $$ 4^x = 4 \Rightarrow x = 1 $$

**Answer:**
$$ x = 1 \quad \text{and} \quad x = \frac{3}{2} $$

---

### Problem (10)

**Equation:**
$$ 5^{x} - 5^{2 - x} = 24 $$

**Solution:**
1. Let $$y = 5^x$$, then $$5^{2 - x} = \frac{25}{y}$$:
   $$ y - \frac{25}{y} = 24 $$
2. Multiply both sides by $$y$$:
   $$ y^2 - 25 = 24y $$
3. Rearrange the equation:
   $$ y^2 - 24y - 25 = 0 $$
4. Solve the quadratic equation:
   $$ y = \frac{24 \pm \sqrt{576 + 100}}{2} = \frac{24 \pm \sqrt{676}}{2} = \frac{24 \pm 26}{2} $$
   $$ y = 25 \quad \text{or} \quad y = -1 $$
5. Since $$y = 5^x > 0$$, discard $$y = -1$$:
   $$ 5^x = 25 \Rightarrow x = 2 $$

**Answer:**
$$ x = 2 $$

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### Problem (12)

**Equation:**
$$ 3^{4x} - 12 \cdot 3^{2x} + 27 = 0 $$

**Solution:**
1. Let $$y = 3^{2x}$$, then $$3^{4x} = y^2$$:
   $$ y^2 - 12y + 27 = 0 $$
2. Solve the quadratic equation:
   $$ y = \frac{12 \pm \sqrt{144 - 108}}{2} = \frac{12 \pm \sqrt{36}}{2} = \frac{12 \pm 6}{2} $$
   $$ y = 9 \quad \text{or} \quad y = 3 $$
3. Convert back to $$x$$:
   - For $$y = 9$$:
     $$ 3^{2x} = 9 \Rightarrow 3^{2x} = 3^2 \Rightarrow 2x = 2 \Rightarrow x = 1 $$
   - For $$y = 3$$:
     $$ 3^{2x} = 3 \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2} $$

**Answer:**
$$ x = \frac{1}{2} \quad \text{and} \quad x = 1 $$

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If you have any further questions or need additional explanations, feel free to ask!
Here are the solutions to the equations: 1. **Problem (2):** $$ x = 1 $$ 2. **Problem (4):** $$ x = -2 $$ 3. **Problem (6):** $$ x = 1 \quad \text{and} \quad x = 3 $$ 4. **Problem (8):** $$ x = 1 \quad \text{and} \quad x = \frac{3}{2} $$ 5. **Problem (10):** $$ x = 2 $$ 6. **Problem (12):** $$ x = \frac{1}{2} \quad \text{and} \quad x = 1 $$ If you need more detailed steps or explanations for any of these solutions, feel free to ask!

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