(c) $$ 4^{x}=8 $$ (f) $$ 3^{x+1}=\frac{\sqrt{3}}{3} $$ (i) $$ 8^{x} \cdot 16^{x-1}=1 $$ (1) $$ (0,375)^{x}=\frac{9}{64} $$

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Sure, let's solve each of the given exponential equations for $$ x $$.

---

### (c) $$ 4^{x} = 8 $$

**Step 1:** Express both sides with base 2.

$$
4 = 2^2 \quad \text{and} \quad 8 = 2^3
$$

**Step 2:** Rewrite the equation using base 2.

$$
(2^2)^x = 2^3 \implies 2^{2x} = 2^3
$$

**Step 3:** Since the bases are equal, set the exponents equal.

$$
2x = 3 \implies x = \frac{3}{2}
$$

**Answer:** $$ x = \frac{3}{2} $$

---

### (f) $$ 3^{x+1} = \frac{\sqrt{3}}{3} $$

**Step 1:** Express $$\frac{\sqrt{3}}{3}$$ using exponents.

$$
\sqrt{3} = 3^{1/2} \quad \text{and} \quad \frac{\sqrt{3}}{3} = 3^{1/2} \cdot 3^{-1} = 3^{-1/2}
$$

**Step 2:** Rewrite the equation.

$$
3^{x+1} = 3^{-1/2}
$$

**Step 3:** Since the bases are equal, set the exponents equal.

$$
x + 1 = -\frac{1}{2} \implies x = -\frac{3}{2}
$$

**Answer:** $$ x = -\frac{3}{2} $$

---

### (i) $$ 8^{x} \cdot 16^{x-1} = 1 $$

**Step 1:** Express 8 and 16 as powers of 2.

$$
8 = 2^3 \quad \text{and} \quad 16 = 2^4
$$

**Step 2:** Rewrite the equation using base 2.

$$
(2^3)^x \cdot (2^4)^{x-1} = 1 \implies 2^{3x} \cdot 2^{4x - 4} = 2^0
$$

**Step 3:** Combine the exponents.

$$
2^{7x - 4} = 2^0
$$

**Step 4:** Set the exponents equal.

$$
7x - 4 = 0 \implies 7x = 4 \implies x = \frac{4}{7}
$$

**Answer:** $$ x = \frac{4}{7} $$

---

### (l) $$ (0.375)^{x} = \frac{9}{64} $$

**Step 1:** Convert 0.375 to a fraction.

$$
0.375 = \frac{3}{8}
$$

**Step 2:** Express $$\frac{9}{64}$$ using exponents.

$$
\frac{9}{64} = \frac{3^2}{8^2} = \left(\frac{3}{8}\right)^2
$$

**Step 3:** Rewrite the equation.

$$
\left(\frac{3}{8}\right)^x = \left(\frac{3}{8}\right)^2
$$

**Step 4:** Since the bases are equal, set the exponents equal.

$$
x = 2
$$

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

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**Summary of Solutions:**

- **(c)** $$ x = \dfrac{3}{2} $$
- **(f)** $$ x = -\dfrac{3}{2} $$
- **(i)** $$ x = \dfrac{4}{7} $$
- **(l)** $$ x = 2 $$
The solutions are: - (c) $$ x = \frac{3}{2} $$ - (f) $$ x = -\frac{3}{2} $$ - (i) $$ x = \frac{4}{7} $$ - (l) $$ x = 2 $$

(c) \( 4^{x}=8 \) (f) \( 3^{x+1}=\frac{\sqrt{3}}{3} \) (i) \( 8^{x} \cdot 16^{x-1}=1 \) (1) \( (0,375)^{x}=\frac{9}{64} \)

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