Solve for $$ x $$ : (a) $$ 4.3^{x}+3^{x}=15 $$ (c) $$ 2^{x+2}+2^{x-2}+2^{x}=84 $$ (e) $$ 3^{2 x}-2.3^{x+2}+81=0 $$ (g) $$ 4^{x}+4.2^{x}-5=0 $$

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Let's solve each equation for $$ x $$ step by step.

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### Problem (a): $$ 4 \cdot 3^{x} + 3^{x} = 15 $$

**Solution:**

1. Factor out $$ 3^x $$:
   $$
   (4 + 1) \cdot 3^{x} = 15 \implies 5 \cdot 3^{x} = 15
   $$

2. Divide both sides by 5:
   $$
   3^{x} = 3
   $$

3. Express 3 as $$ 3^1 $$:
   $$
   3^{x} = 3^1 \implies x = 1
   $$

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

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### Problem (c): $$ 2^{x+2} + 2^{x-2} + 2^{x} = 84 $$

**Solution:**

1. Rewrite each term using properties of exponents:
   $$
   2^{x+2} = 4 \cdot 2^{x}, \quad 2^{x-2} = \frac{1}{4} \cdot 2^{x}, \quad 2^{x} = 2^{x}
   $$

2. Substitute back into the equation:
   $$
   4 \cdot 2^{x} + \frac{1}{4} \cdot 2^{x} + 2^{x} = 84
   $$

3. Factor out $$ 2^{x} $$:
   $$
   \left(4 + \frac{1}{4} + 1\right) \cdot 2^{x} = 84 \implies \frac{21}{4} \cdot 2^{x} = 84
   $$

4. Solve for $$ 2^{x} $$:
   $$
   2^{x} = 84 \cdot \frac{4}{21} = 16
   $$

5. Express 16 as $$ 2^4 $$:
   $$
   2^{x} = 2^{4} \implies x = 4
   $$

**Answer:** $$ x = 4 $$

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### Problem (e): $$ 3^{2x} - 2 \cdot 3^{x+2} + 81 = 0 $$

**Solution:**

1. Let $$ y = 3^{x} $$. Then $$ 3^{2x} = y^2 $$ and $$ 3^{x+2} = 9y $$.

2. Substitute into the equation:
   $$
   y^2 - 2 \cdot 9y + 81 = 0 \implies y^2 - 18y + 81 = 0
   $$

3. Solve the quadratic equation:
   $$
   y = \frac{18 \pm \sqrt{(18)^2 - 4 \cdot 1 \cdot 81}}{2} = \frac{18 \pm 0}{2} = 9
   $$

4. Substitute back $$ y = 3^{x} $$:
   $$
   3^{x} = 9 \implies 3^{x} = 3^2 \implies x = 2
   $$

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

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### Problem (g): $$ 4^{x} + 4 \cdot 2^{x} - 5 = 0 $$

**Solution:**

1. Express $$ 4^{x} $$ in terms of $$ 2^{x} $$:
   $$
   4^{x} = (2^2)^{x} = 2^{2x}
   $$

2. Let $$ y = 2^{x} $$. Then $$ 2^{2x} = y^2 $$.

3. Substitute into the equation:
   $$
   y^2 + 4y - 5 = 0
   $$

4. Solve the quadratic equation:
   $$
   y = \frac{-4 \pm \sqrt{16 + 20}}{2} = \frac{-4 \pm 6}{2}
   $$
   $$
   y = 1 \quad \text{(since $$ y = 2^x > 0 $$)}
   $$

5. Substitute back $$ y = 2^{x} $$:
   $$
   2^{x} = 1 \implies x = 0
   $$

**Answer:** $$ x = 0 $$

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### Summary of Solutions:
- **(a)** $$ x = 1 $$
- **(c)** $$ x = 4 $$
- **(e)** $$ x = 2 $$
- **(g)** $$ x = 0 $$
The solutions are: - (a) $$ x = 1 $$ - (c) $$ x = 4 $$ - (e) $$ x = 2 $$ - (g) $$ x = 0 $$

Solve for \( x \) : (a) \( 4.3^{x}+3^{x}=15 \) (c) \( 2^{x+2}+2^{x-2}+2^{x}=84 \) (e) \( 3^{2 x}-2.3^{x+2}+81=0 \) (g) \( 4^{x}+4.2^{x}-5=0 \)

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