Los op vir \( x \) : \( 1 \quad 8^{x} \cdot \frac{1}{4^{x-1}}=\frac{1}{32} \) \( 2 \quad 3^{x+1}=9^{x} \cdot \frac{1}{27} \) \( 3 \quad 3^{x^{2}+2}=9 \cdot \frac{1}{27^{x}} \) \( 4 \quad 5^{x^{2}-1} \cdot \frac{1}{25}=25^{x} \) \( 5 \quad 2^{2 x} \cdot 4^{x+1}=1 \) \( 6 \quad 12^{x} \cdot 12^{x}=144 \) \( 7 \quad 2^{x^{2}} \cdot 2^{x^{2}-2}+1=65 \) \( 8 \quad \frac{4^{x+1} \times\left(\frac{1}{4}\right)^{-x}}{2^{2 x+1}}=\frac{1}{32} \) \( 9 \quad 81.9^{x-1}=\frac{1}{27} \) \( 10 \quad 10^{2 x^{2}} \cdot 10^{x+1}=10000 \)

Solve the equation by following steps: - step0: Solve for \(x\): \(12^{x}\times 12^{x}=144\) - step1: Multiply the terms: \(12^{2x}=144\) - step2: Rewrite in exponential form: \(12^{2x}=12^{2}\) - step3: Set the exponents equal: \(2x=2\) - step4: Divide both sides: \(\frac{2x}{2}=\frac{2}{2}\) - step5: Divide the numbers: \(x=1\) Solve the equation \( 8^{x} \cdot \frac{1}{4^{x-1}}=\frac{1}{32} \). Solve the equation by following steps: - step0: Solve for \(x\): \(8^{x}\times \frac{1}{4^{x-1}}=\frac{1}{32}\) - step1: Simplify: \(2^{x+2}=\frac{1}{32}\) - step2: Rewrite in exponential form: \(2^{x+2}=2^{-5}\) - step3: Set the exponents equal: \(x+2=-5\) - step4: Move the constant to the right side: \(x=-5-2\) - step5: Subtract the numbers: \(x=-7\) Solve the equation \( 3^{x+1}=9^{x} \cdot \frac{1}{27} \). Solve the equation by following steps: - step0: Solve for \(x\): \(3^{x+1}=9^{x}\times \frac{1}{27}\) - step1: Multiply the terms: \(3^{x+1}=3^{2x-3}\) - step2: Set the exponents equal: \(x+1=2x-3\) - step3: Move the expression to the left side: \(x-2x=-3-1\) - step4: Add and subtract: \(-x=-3-1\) - step5: Add and subtract: \(-x=-4\) - step6: Change the signs: \(x=4\) Solve the equation \( 2^{2 x} \cdot 4^{x+1}=1 \). Solve the equation by following steps: - step0: Solve for \(x\): \(2^{2x}\times 4^{x+1}=1\) - step1: Multiply the terms: \(2^{4x+2}=1\) - step2: Rewrite in exponential form: \(2^{4x+2}=2^{0}\) - step3: Set the exponents equal: \(4x+2=0\) - step4: Move the constant to the right side: \(4x=0-2\) - step5: Remove 0: \(4x=-2\) - step6: Divide both sides: \(\frac{4x}{4}=\frac{-2}{4}\) - step7: Divide the numbers: \(x=-\frac{1}{2}\) Solve the equation \( 3^{x^{2}+2}=9 \cdot \frac{1}{27^{x}} \). Solve the equation by following steps: - step0: Solve for \(x\): \(3^{x^{2}+2}=9\times \frac{1}{27^{x}}\) - step1: Simplify: \(3^{x^{2}+2}=3^{-3x+2}\) - step2: Set the exponents equal: \(x^{2}+2=-3x+2\) - step3: Cancel equal terms: \(x^{2}=-3x\) - step4: Move the expression to the left side: \(x^{2}-\left(-3x\right)=0\) - step5: Add or subtract both sides: \(x^{2}+3x=0\) - step6: Factor the expression: \(x\left(x+3\right)=0\) - step7: Separate into possible cases: \(\begin{align}&x=0\\&x+3=0\end{align}\) - step8: Solve the equation: \(\begin{align}&x=0\\&x=-3\end{align}\) - step9: Rewrite: \(x_{1}=-3,x_{2}=0\) Solve the equation \( \frac{4^{x+1} \times\left(\frac{1}{4}\right)^{-x}}{2^{2 x+1}}=\frac{1}{32} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{4^{x+1}\left(\frac{1}{4}\right)^{-x}}{2^{2x+1}}=\frac{1}{32}\) - step1: Reduce the fraction: \(2^{2x+1}=\frac{1}{32}\) - step2: Rewrite in exponential form: \(2^{2x+1}=2^{-5}\) - step3: Set the exponents equal: \(2x+1=-5\) - step4: Move the constant to the right side: \(2x=-5-1\) - step5: Subtract the numbers: \(2x=-6\) - step6: Divide both sides: \(\frac{2x}{2}=\frac{-6}{2}\) - step7: Divide the numbers: \(x=-3\) Solve the equation \( 2^{x^{2}} \cdot 2^{x^{2}-2}+1=65 \). Solve the equation by following steps: - step0: Solve for \(x\): \(2^{\left(x^{2}\right)}\times 2^{x^{2}-2}+1=65\) - step1: Multiply the terms: \(2^{2x^{2}-2}+1=65\) - step2: Move the constant to the right side: \(2^{2x^{2}-2}=65-1\) - step3: Subtract the numbers: \(2^{2x^{2}-2}=64\) - step4: Rewrite in exponential form: \(2^{2x^{2}-2}=2^{6}\) - step5: Set the exponents equal: \(2x^{2}-2=6\) - step6: Move the constant to the right side: \(2x^{2}=6+2\) - step7: Add the numbers: \(2x^{2}=8\) - step8: Divide both sides: \(\frac{2x^{2}}{2}=\frac{8}{2}\) - step9: Divide the numbers: \(x^{2}=4\) - step10: Simplify the expression: \(x=\pm \sqrt{4}\) - step11: Simplify: \(x=\pm 2\) - step12: Separate into possible cases: \(\begin{align}&x=2\\&x=-2\end{align}\) - step13: Rewrite: \(x_{1}=-2,x_{2}=2\) Solve the equation \( 5^{x^{2}-1} \cdot \frac{1}{25}=25^{x} \). Solve the equation by following steps: - step0: Solve for \(x\): \(5^{x^{2}-1}\times \frac{1}{25}=25^{x}\) - step1: Multiply the terms: \(5^{x^{2}-3}=25^{x}\) - step2: Rewrite the expression: \(5^{x^{2}-3}=5^{2x}\) - step3: Set the exponents equal: \(x^{2}-3=2x\) - step4: Move the expression to the left side: \(x^{2}-3-2x=0\) - step5: Factor the expression: \(\left(x-3\right)\left(x+1\right)=0\) - step6: Separate into possible cases: \(\begin{align}&x-3=0\\&x+1=0\end{align}\) - step7: Solve the equation: \(\begin{align}&x=3\\&x=-1\end{align}\) - step8: Rewrite: \(x_{1}=-1,x_{2}=3\) Solve the equation \( 81.9^{x-1}=\frac{1}{27} \). Solve the equation by following steps: - step0: Solve for \(x\): \(81.9^{x-1}=\frac{1}{27}\) - step1: Convert the expressions: \(\left(\frac{819}{10}\right)^{x-1}=\frac{1}{27}\) - step2: Take the logarithm of both sides: \(\log_{\frac{819}{10}}{\left(\left(\frac{819}{10}\right)^{x-1}\right)}=\log_{\frac{819}{10}}{\left(\frac{1}{27}\right)}\) - step3: Evaluate the logarithm: \(x-1=\log_{\frac{819}{10}}{\left(\frac{1}{27}\right)}\) - step4: Simplify: \(x-1=-3\log_{\frac{819}{10}}{\left(3\right)}\) - step5: Move the constant to the right side: \(x=-3\log_{\frac{819}{10}}{\left(3\right)}+1\) Solve the equation \( 10^{2 x^{2}} \cdot 10^{x+1}=10000 \). Solve the equation by following steps: - step0: Solve for \(x\): \(10^{2x^{2}}\times 10^{x+1}=10000\) - step1: Multiply: \(10^{2x^{2}+x+1}=10000\) - step2: Rewrite in exponential form: \(10^{2x^{2}+x+1}=10^{4}\) - step3: Set the exponents equal: \(2x^{2}+x+1=4\) - step4: Move the expression to the left side: \(2x^{2}+x+1-4=0\) - step5: Subtract the numbers: \(2x^{2}+x-3=0\) - step6: Factor the expression: \(\left(x-1\right)\left(2x+3\right)=0\) - step7: Separate into possible cases: \(\begin{align}&x-1=0\\&2x+3=0\end{align}\) - step8: Solve the equation: \(\begin{align}&x=1\\&x=-\frac{3}{2}\end{align}\) - step9: Rewrite: \(x_{1}=-\frac{3}{2},x_{2}=1\) Here are the solutions for each equation step by step: 1. **Equation:** \( 8^{x} \cdot \frac{1}{4^{x-1}}=\frac{1}{32} \) **Solution:** \[ 8^{x} \cdot 4^{-(x-1)} = \frac{1}{32} \] Converting to base 2: \[ (2^3)^{x} \cdot (2^2)^{-(x-1)} = 2^{-5} \] Simplifying: \[ 2^{3x} \cdot 2^{-2(x-1)} = 2^{-5} \] \[ 2^{3x - 2x + 2} = 2^{-5} \] \[ 2^{x + 2} = 2^{-5} \] Setting exponents equal: \[ x + 2 = -5 \implies x = -7 \] 2. **Equation:** \( 3^{x+1}=9^{x} \cdot \frac{1}{27} \) **Solution:** \[ 3^{x+1} = (3^2)^{x} \cdot \frac{1}{3^3} \] \[ 3^{x+1} = 3^{2x - 3} \] Setting exponents equal: \[ x + 1 = 2x - 3 \implies x = 4 \] 3. **Equation:** \( 3^{x^{2}+2}=9 \cdot \frac{1}{27^{x}} \) **Solution:** \[ 3^{x^{2}+2} = 3^2 \cdot 3^{-3x} \] \[ 3^{x^{2}+2} = 3^{2 - 3x} \] Setting exponents equal: \[ x^{2} + 2 = 2 - 3x \] Rearranging: \[ x^{2} + 3x = 0 \implies x(x + 3) = 0 \] Solutions: \( x_{1} = 0, x_{2} = -3 \) 4. **Equation:** \( 5^{x^{2}-1} \cdot \frac{1}{25}=25^{x} \) **Solution:** \[ 5^{x^{2}-1} \cdot 5^{-2} = 5^{2x} \] \[ 5^{x^{2}-3} = 5^{2x} \] Setting exponents equal: \[ x^{2} - 3 = 2x \implies x^{2} - 2x - 3 = 0 \] Factoring: \[ (x - 3)(x + 1) = 0 \implies x_{1} = 3, x_{2} = -1 \] 5. **Equation:** \( 2^{2 x} \cdot 4^{x+1}=1 \) **Solution:** \[ 2^{2x} \cdot (2^2)^{x+1} = 1 \] \[ 2^{2x + 2(x+1)} = 1 \implies 2^{2x + 2x + 2} = 1 \] \[ 2^{4x + 2} = 2^0 \] Setting exponents equal: \[ 4x + 2 = 0 \implies x = -\frac{1}{2} \] 6. **Equation:** \( 12^{x} \cdot 12^{x}=144 \) **Solution:** \[ 12^{2x} = 12^2 \] Setting exponents equal: \[ 2x = 2 \implies x = 1 \] 7. **Equation:** \( 2^{x^{2}} \cdot 2^{x^{2}-2}+1=65 \) **Solution:** \[ 2^{2x^{2}-2} + 1 = 65 \] \[ 2^{2x^{2}-2} = 64 \implies 2^{2x^{2}-2} = 2^6 \] Setting exponents equal: \[ 2x^{2} - 2 = 6 \implies 2x^{2} = 8 \implies x^{2} = 4 \] Solutions: \( x_{1} = 2, x_{2} = -2 \) 8. **Equation:** \( \frac{4^{x+1} \times\left(\frac{1}{4}\right)^{-x}}{2^{2 x+1}}=\frac{1}{32} \) **Solution:** \[ \frac{(2^2)^{x+1} \cdot (2^{-2})^{-x}}{2^{2x+1}} = 2^{-5} \] \[ \frac{2^{2x + 2} \cdot 2^{2x}}{2^{2x + 1}} = 2^{-5} \] \[ 2^{2x + 2 + 2x - (2x + 1)} = 2^{-5} \] \[ 2^{2x + 1} = 2^{-5} \] Setting exponents equal: \[ 2x + 1 = -5 \implies x = -3 \] 9. **Equation:** \( 81.9^{x-1}=\frac{1}{27} \) **Solution:** \[ 81.9^{x-1} = 27^{-1} \] Converting to base 3: \[ (3^4)^{x-1} = (3^3)^{-1} \] \[ 3^{4(x-1)} = 3^{-3} \] Setting exponents equal: \[ 4(x-1) = -3 \implies 4x - 4 = -3 \implies 4x = 1 \implies x = \frac{1}{4} \] 10. **Equation