1) \( 3^{5 x-1} 8^{2 x+1} \) \( 27^{4 x+1} \) 2) \( \frac{3^{x-1} \cdot 45^{3 x-1} 25^{2-2}}{15^{x+1} \times 27^{22}} \)

Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{3^{x-1}\times 45^{3x-1}\times 25^{2-2}}{15^{x+1}\times 27^{22}}=0\) - step1: Simplify: \(\frac{5^{2x-2}}{9^{35-3x}}=0\) - step2: The statement is false for any value of \(x:\) \(x \notin \mathbb{R}\) Solve the equation \( 3^{5 x-1} 8^{2 x+1} = 27^{4 x+1} \). Solve the equation by following steps: - step0: Solve for \(x\): \(3^{5x-1}\times 8^{2x+1}=27^{4x+1}\) - step1: Take the logarithm of both sides: \(\ln{\left(3^{5x-1}\times 8^{2x+1}\right)}=\ln{\left(27^{4x+1}\right)}\) - step2: Simplify: \(5\ln{\left(3\right)}\times x-\ln{\left(3\right)}+6\ln{\left(2\right)}\times x+3\ln{\left(2\right)}=3\left(4x+1\right)\times \ln{\left(3\right)}\) - step3: Collect like terms: \(\left(5\ln{\left(3\right)}+6\ln{\left(2\right)}\right)x-\ln{\left(3\right)}+3\ln{\left(2\right)}=3\left(4x+1\right)\times \ln{\left(3\right)}\) - step4: Calculate: \(\left(5\ln{\left(3\right)}+6\ln{\left(2\right)}\right)x-\ln{\left(3\right)}+3\ln{\left(2\right)}=3\ln{\left(3\right)}\times \left(4x+1\right)\) - step5: Expand the expression: \(5\ln{\left(3\right)}\times x+6\ln{\left(2\right)}\times x-\ln{\left(3\right)}+3\ln{\left(2\right)}=3\ln{\left(3\right)}\times \left(4x+1\right)\) - step6: Expand the expression: \(5\ln{\left(3\right)}\times x+6\ln{\left(2\right)}\times x-\ln{\left(3\right)}+3\ln{\left(2\right)}=12\ln{\left(3\right)}\times x+3\ln{\left(3\right)}\) - step7: Collect like terms: \(\left(5\ln{\left(3\right)}+6\ln{\left(2\right)}\right)x-\ln{\left(3\right)}+3\ln{\left(2\right)}=12\ln{\left(3\right)}\times x+3\ln{\left(3\right)}\) - step8: Move the expression to the left side: \(\left(5\ln{\left(3\right)}+6\ln{\left(2\right)}\right)x-12\ln{\left(3\right)}\times x=3\ln{\left(3\right)}+\ln{\left(3\right)}-3\ln{\left(2\right)}\) - step9: Add and subtract: \(\left(-7\ln{\left(3\right)}+6\ln{\left(2\right)}\right)x=3\ln{\left(3\right)}+\ln{\left(3\right)}-3\ln{\left(2\right)}\) - step10: Add and subtract: \(\left(-7\ln{\left(3\right)}+6\ln{\left(2\right)}\right)x=4\ln{\left(3\right)}-3\ln{\left(2\right)}\) - step11: Change the signs: \(\left(7\ln{\left(3\right)}-6\ln{\left(2\right)}\right)x=-4\ln{\left(3\right)}+3\ln{\left(2\right)}\) - step12: Divide both sides: \(\frac{\left(7\ln{\left(3\right)}-6\ln{\left(2\right)}\right)x}{7\ln{\left(3\right)}-6\ln{\left(2\right)}}=\frac{-4\ln{\left(3\right)}+3\ln{\left(2\right)}}{7\ln{\left(3\right)}-6\ln{\left(2\right)}}\) - step13: Divide the numbers: \(x=\frac{-4\ln{\left(3\right)}+3\ln{\left(2\right)}}{7\ln{\left(3\right)}-6\ln{\left(2\right)}}\) - step14: Simplify: \(x=\frac{\ln{\left(\frac{2^{3}}{3^{4}}\right)}}{\ln{\left(\frac{3^{7}}{2^{6}}\right)}}\) - step15: Evaluate: \(x=\frac{3\ln{\left(2\right)}-4\ln{\left(3\right)}}{7\ln{\left(3\right)}-6\ln{\left(2\right)}}\) Let's solve the two problems step by step. ### Problem 1: Solve \( 3^{5x-1} \cdot 8^{2x+1} = 27^{4x+1} \) 1. **Rewrite the bases in terms of powers of 3:** - \( 3^{5x-1} \) remains as is. - \( 8 = 2^3 \), so \( 8^{2x+1} = (2^3)^{2x+1} = 2^{6x+3} \). - \( 27 = 3^3 \), so \( 27^{4x+1} = (3^3)^{4x+1} = 3^{12x+3} \). 2. **Substituting back into the equation:** \[ 3^{5x-1} \cdot 2^{6x+3} = 3^{12x+3} \] 3. **Equate the powers of 3:** \[ 5x - 1 = 12x + 3 \] 4. **Solve for \( x \):** \[ 5x - 12x = 3 + 1 \implies -7x = 4 \implies x = -\frac{4}{7} \] 5. **Final result:** \[ x = \frac{3\ln{(2)} - 4\ln{(3)}}{7\ln{(3)} - 6\ln{(2)}} \] ### Problem 2: Solve \( \frac{3^{x-1} \cdot 45^{3x-1} \cdot 25^{2-2}}{15^{x+1} \cdot 27^{22}} = 0 \) 1. **Simplify the expression:** - \( 25^{2-2} = 25^0 = 1 \), so it can be ignored. - Rewrite \( 45 \) and \( 15 \) in terms of their prime factors: - \( 45 = 3^2 \cdot 5 \) - \( 15 = 3 \cdot 5 \) - Thus, \( 45^{3x-1} = (3^2 \cdot 5)^{3x-1} = 3^{6x-2} \cdot 5^{3x-1} \) - And \( 15^{x+1} = (3 \cdot 5)^{x+1} = 3^{x+1} \cdot 5^{x+1} \) - \( 27 = 3^3 \), so \( 27^{22} = 3^{66} \) 2. **Substituting back into the equation:** \[ \frac{3^{x-1} \cdot 3^{6x-2} \cdot 5^{3x-1}}{3^{x+1} \cdot 5^{x+1} \cdot 3^{66}} = 0 \] 3. **Combine the powers of 3 and 5:** \[ \frac{3^{(x-1) + (6x-2)}}{3^{(x+1) + 66}} \cdot \frac{5^{(3x-1)}}{5^{(x+1)}} = 0 \] This simplifies to: \[ \frac{3^{7x-3}}{3^{x+67}} \cdot \frac{5^{3x-1}}{5^{x+1}} = 0 \] 4. **This implies:** \[ 3^{7x-3 - (x+67)} \cdot 5^{3x-1 - (x+1)} = 0 \] Which simplifies to: \[ 3^{6x - 70} \cdot 5^{2x - 2} = 0 \] 5. **Since the product of two positive numbers cannot be zero, we conclude:** \[ x \notin \mathbb{R} \quad \text{(No real solution)} \] ### Summary of Results: 1. For the first problem, the solution is: \[ x = \frac{3\ln{(2)} - 4\ln{(3)}}{7\ln{(3)} - 6\ln{(2)}} \] 2. For the second problem, there is no real solution: \[ x \notin \mathbb{R} \quad \text{(No real solution)} \]