i \( 3^{x}-3^{x-2}=24 \) if \( 3^{x}+3^{x+2}=10 \) iii \( 2^{x}+2^{x}+2^{x}=96 \) iv \( 12^{x}+12^{x}=288 \) \( 2^{x}+4^{x+1}=80 \)

Sure, let's solve each of the given exponential equations step by step to find the value of \( x \). --- ### **i. \( 3^{x} - 3^{x-2} = 24 \)** **Step 1:** Factor out \( 3^{x-2} \) \[ 3^{x} - 3^{x-2} = 3^{x-2} \left( 3^{2} - 1 \right) = 3^{x-2} \times (9 - 1) = 3^{x-2} \times 8 \] So, \[ 3^{x-2} \times 8 = 24 \] **Step 2:** Solve for \( 3^{x-2} \) \[ 3^{x-2} = \frac{24}{8} = 3 \] **Step 3:** Express 3 as a power of 3 \[ 3 = 3^{1} \] So, \[ 3^{x-2} = 3^{1} \implies x - 2 = 1 \implies x = 3 \] **Answer:** \( x = 3 \) --- ### **ii. \( 3^{x} + 3^{x+2} = 10 \)** **Step 1:** Factor out \( 3^{x} \) \[ 3^{x} + 3^{x+2} = 3^{x} \left(1 + 3^{2}\right) = 3^{x} \times (1 + 9) = 3^{x} \times 10 \] So, \[ 3^{x} \times 10 = 10 \] **Step 2:** Solve for \( 3^{x} \) \[ 3^{x} = \frac{10}{10} = 1 \] **Step 3:** Express 1 as a power of 3 \[ 1 = 3^{0} \] So, \[ 3^{x} = 3^{0} \implies x = 0 \] **Answer:** \( x = 0 \) --- ### **iii. \( 2^{x} + 2^{x} + 2^{x} = 96 \)** **Step 1:** Combine like terms \[ 3 \times 2^{x} = 96 \] **Step 2:** Solve for \( 2^{x} \) \[ 2^{x} = \frac{96}{3} = 32 \] **Step 3:** Express 32 as a power of 2 \[ 32 = 2^{5} \] So, \[ 2^{x} = 2^{5} \implies x = 5 \] **Answer:** \( x = 5 \) --- ### **iv. \( 12^{x} + 12^{x} = 288 \)** **Step 1:** Combine like terms \[ 2 \times 12^{x} = 288 \] **Step 2:** Solve for \( 12^{x} \) \[ 12^{x} = \frac{288}{2} = 144 \] **Step 3:** Express 144 as a power of 12 Note that \( 12^{2} = 144 \). So, \[ 12^{x} = 12^{2} \implies x = 2 \] **Answer:** \( x = 2 \) --- ### **v. \( 2^{x} + 4^{x+1} = 80 \)** **Step 1:** Express \( 4^{x+1} \) in terms of base 2. Since \( 4 = 2^{2} \), \[ 4^{x+1} = (2^{2})^{x+1} = 2^{2(x+1)} = 2^{2x + 2} \] **Step 2:** Rewrite the equation \[ 2^{x} + 2^{2x + 2} = 80 \] **Step 3:** Let’s set \( y = 2^{x} \). Then \( 2^{2x} = y^{2} \). So, \[ y + y^{2} \times 2^{2} = 80 \implies y + 4y^{2} = 80 \] **Step 4:** Rearrange the equation \[ 4y^{2} + y - 80 = 0 \] **Step 5:** Solve the quadratic equation Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \), where \( a = 4 \), \( b = 1 \), and \( c = -80 \), \[ y = \frac{-1 \pm \sqrt{1 + 1280}}{8} = \frac{-1 \pm \sqrt{1281}}{8} \] Calculate \( \sqrt{1281} \approx 35.8 \), \[ y \approx \frac{-1 + 35.8}{8} \approx \frac{34.8}{8} \approx 4.35 \] We discard the negative solution since \( y = 2^{x} > 0 \). **Step 6:** Solve for \( x \) \[ 2^{x} \approx 4.35 \] Take the natural logarithm on both sides: \[ x \ln 2 \approx \ln 4.35 \implies x \approx \frac{\ln 4.35}{\ln 2} \approx \frac{1.470}{0.693} \approx 2.12 \] **Answer:** \( x \approx 2.12 \) --- **Summary of Solutions:** 1. **i.** \( x = 3 \) 2. **ii.** \( x = 0 \) 3. **iii.** \( x = 5 \) 4. **iv.** \( x = 2 \) 5. **v.** \( x \approx 2.12 \)