Find the value of x in each of the following .
(i) `root5(5x + 2) = 2` (ii) `root3(3x - 2) =4`
(iii) `((3)/(4))^(3)((4)/(3))^(-7)= ((3)/(4))^(2x)` (iv) `5^(x-3) xx 3^(2x -8) = 225`
(v) `(3^(3x) . 3^(2x))/(3^(x)) = root4(3^(20))`

Answer

Let's solve each part of the question step by step.

### (i) Solve for \( x \) in \( \sqrt[5]{5x + 2} = 2 \)

1. **Rewrite the equation**:
\[ (5x + 2)^{\frac{1}{5}} = 2 \]

2. **Raise both sides to the power of 5**:
\[ 5x + 2 = 2^5 \]

3. **Calculate \( 2^5 \)**:
\[ 2^5 = 32 \]
So, we have:
\[ 5x + 2 = 32 \]

4. **Subtract 2 from both sides**:
\[ 5x = 32 - 2 \]
\[ 5x = 30 \]

5. **Divide by 5**:
\[ x = \frac{30}{5} = 6 \]

### (ii) Solve for \( x \) in \( \sqrt[3]{3x - 2} = 4 \)

1. **Rewrite the equation**:
\[ (3x - 2)^{\frac{1}{3}} = 4 \]

2. **Raise both sides to the power of 3**:
\[ 3x - 2 = 4^3 \]

3. **Calculate \( 4^3 \)**:
\[ 4^3 = 64 \]
So, we have:
\[ 3x - 2 = 64 \]

4. **Add 2 to both sides**:
\[ 3x = 64 + 2 \]
\[ 3x = 66 \]

5. **Divide by 3**:
\[ x = \frac{66}{3} = 22 \]

### (iii) Solve for \( x \) in \( \left(\frac{3}{4}\right)^3 \left(\frac{4}{3}\right)^{-7} = \left(\frac{3}{4}\right)^{2x} \)

1. **Rewrite \( \left(\frac{4}{3}\right)^{-7} \)**:
\[ \left(\frac{4}{3}\right)^{-7} = \frac{1}{\left(\frac{4}{3}\right)^7} = \left(\frac{3}{4}\right)^7 \]

2. **Combine the bases**:
\[ \left(\frac{3}{4}\right)^3 \cdot \left(\frac{3}{4}\right)^7 = \left(\frac{3}{4}\right)^{3 + 7} = \left(\frac{3}{4}\right)^{10} \]

3. **Set the exponents equal**:
\[ 10 = 2x \]

4. **Divide by 2**:
\[ x = \frac{10}{2} = 5 \]

### (iv) Solve for \( x \) in \( 5^{x-3} \cdot 3^{2x - 8} = 225 \)

1. **Rewrite 225 as a product of primes**:
\[ 225 = 15^2 = (5 \cdot 3)^2 = 5^2 \cdot 3^2 \]

2. **Set the equation**:
\[ 5^{x-3} \cdot 3^{2x - 8} = 5^2 \cdot 3^2 \]

3. **Equate the exponents of 5**:
\[ x - 3 = 2 \quad \Rightarrow \quad x = 5 \]

4. **Equate the exponents of 3**:
\[ 2x - 8 = 2 \quad \Rightarrow \quad 2x = 10 \quad \Rightarrow \quad x = 5 \]

### (v) Solve for \( x \) in \( \frac{3^{3x} \cdot 3^{2x}}{3^x} = \sqrt[4]{3^{20}} \)

1. **Combine the left side**:
\[ \frac{3^{3x + 2x}}{3^x} = \frac{3^{5x}}{3^x} = 3^{5x - x} = 3^{4x} \]

2. **Rewrite the right side**:
\[ \sqrt[4]{3^{20}} = 3^{20/4} = 3^5 \]

3. **Set the exponents equal**:
\[ 4x = 5 \]

4. **Divide by 4**:
\[ x = \frac{5}{4} \]

### Summary of Solutions:
- (i) \( x = 6 \)
- (ii) \( x = 22 \)
- (iii) \( x = 5 \)
- (iv) \( x = 5 \)
- (v) \( x = \frac{5}{4} \)