Question

OEFENING 5 Los op vir \( x \) : \( 1 \quad 18^{x} \cdot \frac{1}{4^{x-1}}=\frac{1}{32} \) \( 2 \quad 3^{x+1}=9^{x} \cdot \frac{1}{27} \) \( 3 \quad 3^{x^{2}+2}=9 \cdot \frac{1}{27^{x}} \) \( 4 \quad 5^{x^{2}-1} \cdot \frac{1}{25}=25^{x} \) \( 5 \quad 2^{2 x} \cdot 4^{x+1}=1 \) \( 6 \quad 12^{x} .12^{x}=144 \) \( 7 \quad 2^{x^{2}} \cdot 2^{x^{2}-2}+1=65 \) \( 8 \quad \frac{4^{x+1} \times\left(\frac{1}{4}\right)^{-x}}{2^{2 x+1}}=\frac{1}{32} \) \( 9 \quad 81.9^{x-1}=\frac{1}{27} \) \( 10 \quad 10^{2 x^{2}} \cdot 10^{x+1}=10000 \)

Ask by Ortiz Ortiz. in South Africa
Feb 03,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

Here are the solutions for each equation: 1. \( x = -7\log_{\frac{9}{2}}{(2)} \) 2. \( x = 4 \) 3. \( x_{1} = -3, \quad x_{2} = 0 \) 4. \( x_{1} = -1, \quad x_{2} = 3 \) 5. \( x = -0.5 \) 6. \( x = 1 \) 7. \( x_{1} = -2, \quad x_{2} = 2 \) 8. \( x = -3 \) 9. \( x = -2.5 \) 10. \( x_{1} = -1.5, \quad x_{2} = 1 \)

Solution

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 \( 3^{x+1}=9^{x} \cdot \frac{1}{27} \). 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 \( 81 \cdot 9^{x-1}=\frac{1}{27} \). Solve the equation by following steps: - step0: Solve for \(x\): \(81\times 9^{x-1}=\frac{1}{27}\) - step1: Multiply the terms: \(9^{x+1}=\frac{1}{27}\) - step2: Rewrite in exponential form: \(3^{2\left(x+1\right)}=3^{-3}\) - step3: Set the exponents equal: \(2\left(x+1\right)=-3\) - step4: Divide both sides: \(\frac{2\left(x+1\right)}{2}=\frac{-3}{2}\) - step5: Divide the numbers: \(x+1=-\frac{3}{2}\) - step6: Move the constant to the right side: \(x=-\frac{3}{2}-1\) - step7: Subtract the numbers: \(x=-\frac{5}{2}\) 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 \( 3^{x^{2}+2}=9 \cdot \frac{1}{27^{x}} \). 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 \( 5^{x^{2}-1} \cdot \frac{1}{25}=25^{x} \). 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 \( 2^{x^{2}} \cdot 2^{x^{2}-2}+1=65 \). 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 \( 10^{2 x^{2}} \cdot 10^{x+1}=10000 \). 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: \(x_{1}=-\frac{3}{2},x_{2}=1\) Solve the equation \( 18^{x} \cdot \frac{1}{4^{x-1}}=\frac{1}{32} \). Solve the equation by following steps: - step0: Solve for \(x\): \(18^{x}\times \frac{1}{4^{x-1}}=\frac{1}{32}\) - step1: Simplify: \(9^{x}\times 2^{-x+2}=\frac{1}{32}\) - step2: Take the logarithm of both sides: \(\ln{\left(9^{x}\times 2^{-x+2}\right)}=\ln{\left(\frac{1}{32}\right)}\) - step3: Evaluate the logarithm: \(2\ln{\left(3\right)}\times x-\ln{\left(2\right)}\times x+2\ln{\left(2\right)}=\ln{\left(\frac{1}{32}\right)}\) - step4: Evaluate the logarithm: \(2\ln{\left(3\right)}\times x-\ln{\left(2\right)}\times x+2\ln{\left(2\right)}=-5\ln{\left(2\right)}\) - step5: Simplify: \(\ln{\left(\frac{3^{2}}{2}\right)}\times x+2\ln{\left(2\right)}=-5\ln{\left(2\right)}\) - step6: Move the constant to the right side: \(\ln{\left(\frac{3^{2}}{2}\right)}\times x=-5\ln{\left(2\right)}-2\ln{\left(2\right)}\) - step7: Subtract the numbers: \(\ln{\left(\frac{3^{2}}{2}\right)}\times x=-7\ln{\left(2\right)}\) - step8: Divide both sides: \(\frac{\ln{\left(\frac{3^{2}}{2}\right)}\times x}{\ln{\left(\frac{3^{2}}{2}\right)}}=\frac{-7\ln{\left(2\right)}}{\ln{\left(\frac{3^{2}}{2}\right)}}\) - step9: Divide the numbers: \(x=-7\log_{\frac{3^{2}}{2}}{\left(2\right)}\) - step10: Simplify: \(x=-7\log_{\frac{9}{2}}{\left(2\right)}\) Here are the solutions for each equation step by step: 1. **Equation:** \( 18^{x} \cdot \frac{1}{4^{x-1}}=\frac{1}{32} \) **Solution:** \[ x = -7\log_{\frac{9}{2}}{(2)} \] 2. **Equation:** \( 3^{x+1}=9^{x} \cdot \frac{1}{27} \) **Solution:** \[ x = 4 \] 3. **Equation:** \( 3^{x^{2}+2}=9 \cdot \frac{1}{27^{x}} \) **Solution:** \[ x_{1} = -3, \quad x_{2} = 0 \] 4. **Equation:** \( 5^{x^{2}-1} \cdot \frac{1}{25}=25^{x} \) **Solution:** \[ x_{1} = -1, \quad x_{2} = 3 \] 5. **Equation:** \( 2^{2 x} \cdot 4^{x+1}=1 \) **Solution:** \[ x = -\frac{1}{2} \quad \text{(or } x = -0.5\text{)} \] 6. **Equation:** \( 12^{x} \cdot 12^{x}=144 \) **Solution:** \[ x = 1 \] 7. **Equation:** \( 2^{x^{2}} \cdot 2^{x^{2}-2}+1=65 \) **Solution:** \[ x_{1} = -2, \quad x_{2} = 2 \] 8. **Equation:** \( \frac{4^{x+1} \times\left(\frac{1}{4}\right)^{-x}}{2^{2 x+1}}=\frac{1}{32} \) **Solution:** \[ x = -3 \] 9. **Equation:** \( 81 \cdot 9^{x-1}=\frac{1}{27} \) **Solution:** \[ x = -\frac{5}{2} \quad \text{(or } x = -2.5\text{)} \] 10. **Equation:** \( 10^{2 x^{2}} \cdot 10^{x+1}=10000 \) **Solution:** \[ x_{1} = -\frac{3}{2}, \quad x_{2} = 1 \quad \text{(or } x_{1} = -1.5, \quad x_{2} = 1\text{)} \] These solutions provide the values of \( x \) for each equation.

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Wist je dat exponentiële vergelijkingen al sinds de oudheid worden bestudeerd? De concepten erachter zijn te vinden in de werken van grote wiskundigen zoals Euclides en Archimedes. Hun fascinatie voor de eigenschappen van getallen en hoe ze zich verhouden tot elkaar legde de basis voor de moderne algebra en de manier waarop we tegenwoordig exponentiële functies begrijpen! Als je dit soort vergelijkingen in het dagelijks leven tegenkomt, is dat geen toeval! Ze verschijnen bijvoorbeeld in situaties zoals de groei van populaties, het afnemen van radioactiviteit, en zelfs bij het begrijpen van rente bij financiële investeringen. Ze helpen ons om complexe situaties te modelleren en voorspellingen te doen op basis van de eigenschappen van getallen, wat echt nuttig kan zijn!

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