Show that: \( \begin{array}{ll}\text { (a) } 5 \cdot 2^{x}-2^{x+2}=2^{x} & \text { (b) } 9^{x}+3^{2 x+1}=43^{x x} \\ \text { (c) } 2^{2 x-1}+4^{x+1}=2^{x x}\left(\frac{9}{2}\right) & \text { (d) } 2^{x+3}-2^{x+2}=2^{x+2} \\ \text { (e) } 2.10^{x}-5^{x+1} \cdot 2^{x}=-3.10^{x} & \text { (f) } 4.3^{1-x}+3^{2-x}=\frac{21}{3^{x}} \\ \text { Simplify the following: } & \end{array} \)

Certainly! Let's simplify each of the given expressions step by step. --- ### **Problem (a)** **Simplify:** \[ 5 \cdot 2^{x} - 2^{x+2} = 2^{x} \] **Solution:** 1. **Express \(2^{x+2}\) in terms of \(2^{x}\):** \[ 2^{x+2} = 2^{x} \cdot 2^{2} = 4 \cdot 2^{x} \] 2. **Substitute back into the original equation:** \[ 5 \cdot 2^{x} - 4 \cdot 2^{x} = 2^{x} \] 3. **Combine like terms:** \[ (5 - 4) \cdot 2^{x} = 2^{x} \\ 1 \cdot 2^{x} = 2^{x} \] 4. **Conclusion:** \[ 2^{x} = 2^{x} \] This holds true for all real numbers \(x\). Hence, the equation is an identity. --- ### **Problem (b)** **Simplify:** \[ 9^{x} + 3^{2x+1} = 4 \cdot 3^{2x} \] **Solution:** *Assuming the right-hand side (RHS) is \(4 \cdot 3^{2x}\) based on context.* 1. **Express \(9^{x}\) in terms of \(3^{2x}\):** \[ 9^{x} = (3^{2})^{x} = 3^{2x} \] 2. **Express \(3^{2x+1}\) using exponent rules:** \[ 3^{2x+1} = 3^{2x} \cdot 3^{1} = 3 \cdot 3^{2x} \] 3. **Substitute back into the original equation:** \[ 3^{2x} + 3 \cdot 3^{2x} = 4 \cdot 3^{2x} \] 4. **Factor \(3^{2x}\) from the left-hand side (LHS):** \[ (1 + 3) \cdot 3^{2x} = 4 \cdot 3^{2x} \] 5. **Simplify:** \[ 4 \cdot 3^{2x} = 4 \cdot 3^{2x} \] This confirms that both sides are equal for all real numbers \(x\). Hence, the equation is an identity. --- ### **Problem (c)** **Simplify:** \[ 2^{2x-1} + 4^{x+1} = 2^{x} \left(\frac{9}{2}\right) \] **Solution:** 1. **Express all terms with base 2:** \[ 4^{x+1} = (2^{2})^{x+1} = 2^{2(x+1)} = 2^{2x + 2} \] 2. **Rewrite the equation:** \[ 2^{2x - 1} + 2^{2x + 2} = 2^{x} \cdot \frac{9}{2} \] 3. **Factor \(2^{2x - 1}\) from the LHS:** \[ 2^{2x - 1} \left(1 + 2^{3}\right) = 2^{x} \cdot \frac{9}{2} \\ 2^{2x - 1} \cdot 9 = 2^{x} \cdot \frac{9}{2} \] 4. **Divide both sides by 9:** \[ 2^{2x - 1} = 2^{x} \cdot \frac{1}{2} \] 5. **Simplify the right-hand side:** \[ 2^{2x - 1} = 2^{x - 1} \] 6. **Equate exponents (since the bases are the same and non-zero):** \[ 2x - 1 = x - 1 \\ 2x - x = -1 + 1 \\ x = 0 \] 7. **Conclusion:** The equation holds true when \(x = 0\). --- ### **Problem (d)** **Simplify:** \[ 2^{x+3} - 2^{x+2} = 2^{x+2} \] **Solution:** 1. **Express both terms with a common factor:** \[ 2^{x+3} = 2^{x} \cdot 2^{3} = 8 \cdot 2^{x} \\ 2^{x+2} = 2^{x} \cdot 2^{2} = 4 \cdot 2^{x} \] 2. **Substitute back into the equation:** \[ 8 \cdot 2^{x} - 4 \cdot 2^{x} = 2^{x+2} \] 3. **Combine like terms:** \[ (8 - 4) \cdot 2^{x} = 4 \cdot 2^{x} \] 4. **Recognize that \(4 \cdot 2^{x} = 2^{x+2}\):** \[ 4 \cdot 2^{x} = 2^{x} \cdot 2^{2} = 2^{x+2} \] 5. **Conclusion:** \[ 2^{x+2} = 2^{x+2} \] The equation holds true for all real numbers \(x\). Hence, it is an identity. --- ### **Problem (e)** **Simplify:** \[ 2 \cdot 10^{x} - 5^{x+1} \cdot 2^{x} = -3 \cdot 10^{x} \] **Solution:** 1. **Express \(5^{x+1} \cdot 2^{x}\) in terms of \(10^{x}\):** \[ 5^{x+1} \cdot 2^{x} = 5 \cdot 5^{x} \cdot 2^{x} = 5 \cdot (5 \cdot 2)^{x} = 5 \cdot 10^{x} \] 2. **Substitute back into the equation:** \[ 2 \cdot 10^{x} - 5 \cdot 10^{x} = -3 \cdot 10^{x} \] 3. **Combine like terms:** \[ (2 - 5) \cdot 10^{x} = -3 \cdot 10^{x} \\ -3 \cdot 10^{x} = -3 \cdot 10^{x} \] 4. **Conclusion:** The equation holds true for all real numbers \(x\). Hence, it is an identity. --- ### **Problem (f)** **Simplify:** \[ 4 \cdot 3^{1-x} + 3^{2-x} = \frac{21}{3^{x}} \] **Solution:** 1. **Express all terms with negative exponents:** \[ 3^{1-x} = 3^{-(x-1)} = \frac{1}{3^{x-1}} = \frac{3}{3^{x}} \\ 3^{2-x} = 3^{-(x-2)} = \frac{1}{3^{x-2}} = \frac{9}{3^{x}} \] 2. **Rewrite the equation using these expressions:** \[ 4 \cdot \frac{3}{3^{x}} + \frac{9}{3^{x}} = \frac{21}{3^{x}} \] 3. **Combine like terms:** \[ \left(4 \cdot 3 + 9\right) \cdot \frac{1}{3^{x}} = \frac{21}{3^{x}} \\ (12 + 9) \cdot \frac{1}{3^{x}} = \frac{21}{3^{x}} \\ 21 \cdot \frac{1}{3^{x}} = \frac{21}{3^{x}} \] 4. **Conclusion:** \[ \frac{21}{3^{x}} = \frac{21}{3^{x}} \] The equation holds true for all real numbers \(x\). Hence, it is an identity. --- ### **Summary** All the given equations simplify to identities that hold true for all real numbers \(x\), except for Problem (c), which holds when \(x = 0\).