Solutions 1. $$ x^{-\frac{3}{2}}=8 $$ $$ \begin{aligned} \left(x^{-\frac{3}{2}}\right)^{-\frac{2}{3}} & =\left(2^{3}\right)^{-\frac{2}{3}} \quad \checkmark\left(\text { raise both sides to the power of }-\frac{2}{3}\right) \ x^{+1} & =2^{-2} \quad(2)^{3 \times-\frac{2}{3}} \ x & =\frac{1}{4} \end{aligned} $$ 2. $$ \sqrt[5]{x^{4}}=256 \quad $$ (change radical to fractional exponent form) $$ \begin{aligned} x^{\frac{4}{5}} & = \pm(2)^{8} \quad \quad \text { (raise both sides of equation to the power of } \frac{5}{4} \text { ) } \ \left(x^{\frac{4}{5}}\right)^{\frac{5}{4}} & = \pm\left(2^{8}\right)^{\frac{5}{4}} \checkmark \ \therefore x & = \pm(2)^{10} \checkmark= \pm 1024 \end{aligned} $$ [7] Solutions 1. $$ \sqrt{3 x+4}-5=0 $$ $$ \sqrt{3 x+4}=5 $$ (isolate the radical) $$ (\sqrt{3 x+4})^{2}=5^{2} \quad $$ (square both sides of the equation) $$ 3 x+4=25 $$ $$ 3 x=21 $$ $$ x=7 $$ 2. $$ \sqrt{3 x-5}-x=5 $$ $$ \begin{aligned} \sqrt{3 x-5} & =x-5 \ (\sqrt{3 x-5})^{2} \Omega & =(x-5)^{2} \ 3 x-5 & =x^{2}-10 x+25 \ 0 & =x^{2}-13 x+30 \ 0 & =(x-10)(x-3) \ x & =10 \text { or } x=3 \end{aligned} $$ Check your answer: If $$ x=10 $$ LHS: $$ \sqrt{3(10)-5}-10 $$ Check: LHS: $$ \sqrt{3(7)+4}-5 $$ $$ =\sqrt{21+4}-5 $$ $$ =\sqrt{25}-5 $$ $$ =0 $$ $$ =\text { RHS } $$ $$ \therefore x=7 $$ is a solution (always isolate the radical first) (square both sides) Remember: $$ (x-5)^{2} \neq x^{2}+25 $$ (quadratic equation, set $$ =0 $$ ) (factorise the trinomial and make each factor $$ =0 $$ )

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Let's go through the solutions step by step, ensuring clarity and correctness in each part.

### Problem 1: Solve $$ x^{-\frac{3}{2}}=8 $$

1. **Raise both sides to the power of $$-\frac{2}{3}$$**:
   $$
   \left(x^{-\frac{3}{2}}\right)^{-\frac{2}{3}} = \left(8\right)^{-\frac{2}{3}}
   $$

2. **Simplify the left side**:
   $$
   x^{1} = 8^{-\frac{2}{3}}
   $$

3. **Calculate $$8^{-\frac{2}{3}}$$**:
   - First, express $$8$$ as $$2^3$$:
   $$
   8^{-\frac{2}{3}} = (2^3)^{-\frac{2}{3}} = 2^{-2} = \frac{1}{4}
   $$

4. **Final result**:
   $$
   x = \frac{1}{4}
   $$

### Problem 2: Solve $$ \sqrt[5]{x^{4}}=256 $$

1. **Change radical to fractional exponent form**:
   $$
   x^{\frac{4}{5}} = 256
   $$

2. **Raise both sides to the power of $$\frac{5}{4}$$**:
   $$
   \left(x^{\frac{4}{5}}\right)^{\frac{5}{4}} = 256^{\frac{5}{4}}
   $$

3. **Simplify the left side**:
   $$
   x = 256^{\frac{5}{4}}
   $$

4. **Calculate $$256$$**:
   - Since $$256 = 2^8$$:
   $$
   256^{\frac{5}{4}} = (2^8)^{\frac{5}{4}} = 2^{10} = 1024
   $$

5. **Final result**:
   $$
   x = 1024
   $$

### Problem 3: Solve $$ \sqrt{3x+4}-5=0 $$

1. **Isolate the radical**:
   $$
   \sqrt{3x+4} = 5
   $$

2. **Square both sides**:
   $$
   (\sqrt{3x+4})^{2} = 5^{2}
   $$

3. **Simplify**:
   $$
   3x + 4 = 25
   $$

4. **Solve for $$x$$**:
   $$
   3x = 21 \implies x = 7
   $$

### Problem 4: Solve $$ \sqrt{3x-5}-x=5 $$

1. **Isolate the radical**:
   $$
   \sqrt{3x-5} = x - 5
   $$

2. **Square both sides**:
   $$
   (\sqrt{3x-5})^{2} = (x-5)^{2}
   $$

3. **Expand both sides**:
   $$
   3x - 5 = x^2 - 10x + 25
   $$

4. **Rearrange to form a quadratic equation**:
   $$
   0 = x^2 - 13x + 30
   $$

5. **Factor the quadratic**:
   $$
   0 = (x - 10)(x - 3)
   $$

6. **Find the solutions**:
   $$
   x = 10 \quad \text{or} \quad x = 3
   $$

### Verification of Solutions

- For $$x = 10$$:
   $$
   \sqrt{3(10)-5} - 10 = \sqrt{30 - 5} - 10 = \sqrt{25} - 10 = 5 - 10 = -5 \quad \text{(not a solution)}
   $$

- For $$x = 3$$:
   $$
   \sqrt{3(3)-5} - 3 = \sqrt{9 - 5} - 3 = \sqrt{4} - 3 = 2 - 3 = -1 \quad \text{(not a solution)}
   $$

### Conclusion

The valid solution for the first three problems is:
1. $$ x = \frac{1}{4} $$
2. $$ x = 1024 $$
3. $$ x = 7 $$

For the fourth problem, both potential solutions $$x = 10$$ and $$x = 3$$ do not satisfy the original equation, indicating that there are no valid solutions for that equation.
Solutions: 1. $$ x = \frac{1}{4} $$ 2. $$ x = 1024 $$ 3. $$ x = 7 $$ 4. No solution

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