Step 1: Express 81 as a power of 3: 81 = 3⁴

Step 2: Rewrite the equation: 3ˣ⁺¹ = 3⁴

Step 3: Set exponents equal: x + 1 = 4

Step 4: Solve for x: x = 3

Step 5: Check: 3³⁺¹ = 3⁴ = 81 ✓

Answer: x = 3

The bases don't match easily, so use logarithms:

$\ln(4^x) = \ln(20)$

Use power property: $x \ln(4) = \ln(20)$

Solve for $x$: $x = \frac{\ln(20)}{\ln(4)} \approx 2.161$

Answer: $x = \frac{\ln(20)}{\ln(4)}$ or approximately $2.161$

Step 1: Express 125 as a power of 5: 125 = 5³

Step 2: Rewrite the equation: 5²ˣ = (5³)ˣ⁻¹

Step 3: Apply power rule (bᵐ)ⁿ = bᵐⁿ: 5²ˣ = 5³⁽ˣ⁻¹⁾ 5²ˣ = 5³ˣ⁻³

Step 4: Set exponents equal: 2x = 3x - 3

Step 5: Solve for x: 2x - 3x = -3 -x = -3 x = 3

Step 6: Check: Left: 5²⁽³⁾ = 5⁶ Right: 125³⁻¹ = 125² = (5³)² = 5⁶ ✓

Answer: x = 3

Step 1: Take logarithm of both sides: We can use any base, but log₁₀ or ln are common log(2ˣ) = log(15)

Step 2: Apply power rule: x log(2) = log(15)

Step 3: Solve for x: x = log(15)/log(2)

Step 4: Calculate (using calculator): log(15) ≈ 1.1761 log(2) ≈ 0.3010 x ≈ 1.1761/0.3010 x ≈ 3.907

Step 5: Check: 2³·⁹⁰⁷ ≈ 15.00 ✓

Answer: x = log(15)/log(2) ≈ 3.907

Step 1: Isolate the exponential $2^{x+1} = 16$

Step 2: Write 16 as a power of 2 $2^{x+1} = 2^4$

Step 3: Set exponents equal $x + 1 = 4$

Step 4: Solve $x = 3$

Check: $3 \cdot 2^{3+1} = 3 \cdot 2^4 = 3 \cdot 16 = 48$ ✓

Answer: $x = 3$

Step 1: Express 4ˣ in terms of 2ˣ: 4ˣ = (2²)ˣ = 2²ˣ = (2ˣ)²

Step 2: Express 2ˣ⁺¹: 2ˣ⁺¹ = 2ˣ · 2¹ = 2 · 2ˣ

Step 3: Let u = 2ˣ, then substitute: (2ˣ)² - 2 · 2ˣ - 8 = 0 u² - 2u - 8 = 0

Step 4: Factor the quadratic: (u - 4)(u + 2) = 0

Step 5: Solve for u: u = 4 or u = -2

Step 6: Substitute back 2ˣ for u: 2ˣ = 4 or 2ˣ = -2

Step 7: Solve each equation: 2ˣ = 4 → 2ˣ = 2² → x = 2 ✓ 2ˣ = -2 → No solution (2ˣ is always positive)

Step 8: Check x = 2: 4² - 2²⁺¹ - 8 = 16 - 8 - 8 = 0 ✓

Answer: x = 2