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

Ask by Knight Cox. in South Africa
Jan 23,2025

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The solutions are: 1. \( x = 3 \) 2. \( x = 0 \) 3. \( x = 5 \) 4. \( x = 2 \) 5. \( x = \log_{2}(-1+\sqrt{1281})-3 \)

Solución

Solve the equation \( 3^{x}-3^{x-2}=24 \). Solve the equation by following steps: - step0: Solve for \(x\): \(3^{x}-3^{x-2}=24\) - step1: Subtract the terms: \(8\times 3^{x-2}=24\) - step2: Divide both sides: \(\frac{8\times 3^{x-2}}{8}=\frac{24}{8}\) - step3: Divide the numbers: \(3^{x-2}=3\) - step4: Set the exponents equal: \(x-2=1\) - step5: Move the constant to the right side: \(x=1+2\) - step6: Add the numbers: \(x=3\) Solve the equation \( 3^{x}+3^{x+2}=10 \). Solve the equation by following steps: - step0: Solve for \(x\): \(3^{x}+3^{x+2}=10\) - step1: Add the terms: \(10\times 3^{x}=10\) - step2: Divide both sides: \(\frac{10\times 3^{x}}{10}=\frac{10}{10}\) - step3: Divide the numbers: \(3^{x}=1\) - step4: Rewrite in exponential form: \(3^{x}=3^{0}\) - step5: Set the exponents equal: \(x=0\) Solve the equation \( 2^{x}+2^{x}+2^{x}=96 \). Solve the equation by following steps: - step0: Solve for \(x\): \(2^{x}+2^{x}+2^{x}=96\) - step1: Simplify: \(3\times 2^{x}=96\) - step2: Divide both sides: \(\frac{3\times 2^{x}}{3}=\frac{96}{3}\) - step3: Divide the numbers: \(2^{x}=32\) - step4: Rewrite in exponential form: \(2^{x}=2^{5}\) - step5: Set the exponents equal: \(x=5\) Solve the equation \( 12^{x}+12^{x}=288 \). Solve the equation by following steps: - step0: Solve for \(x\): \(12^{x}+12^{x}=288\) - step1: Add the terms: \(2\times 12^{x}=288\) - step2: Divide both sides: \(\frac{2\times 12^{x}}{2}=\frac{288}{2}\) - step3: Divide the numbers: \(12^{x}=144\) - step4: Rewrite in exponential form: \(12^{x}=12^{2}\) - step5: Set the exponents equal: \(x=2\) Solve the equation \( 2^{x}+4^{x+1}=80 \). Solve the equation by following steps: - step0: Solve for \(x\): \(2^{x}+4^{x+1}=80\) - step1: Move the expression to the left side: \(2^{x}+4^{x+1}-80=0\) - step2: Rewrite the expression: \(2^{x}+4\times 4^{x}-80=0\) - step3: Rewrite the expression: \(2^{x}+4\times 2^{2x}-80=0\) - step4: Use substitution: \(t+4t^{2}-80=0\) - step5: Rewrite in standard form: \(4t^{2}+t-80=0\) - step6: Solve using the quadratic formula: \(t=\frac{-1\pm \sqrt{1^{2}-4\times 4\left(-80\right)}}{2\times 4}\) - step7: Simplify the expression: \(t=\frac{-1\pm \sqrt{1^{2}-4\times 4\left(-80\right)}}{8}\) - step8: Simplify the expression: \(t=\frac{-1\pm \sqrt{1281}}{8}\) - step9: Separate into possible cases: \(\begin{align}&t=\frac{-1+\sqrt{1281}}{8}\\&t=\frac{-1-\sqrt{1281}}{8}\end{align}\) - step10: Rewrite the fraction: \(\begin{align}&t=\frac{-1+\sqrt{1281}}{8}\\&t=-\frac{1+\sqrt{1281}}{8}\end{align}\) - step11: Substitute back: \(\begin{align}&2^{x}=\frac{-1+\sqrt{1281}}{8}\\&2^{x}=-\frac{1+\sqrt{1281}}{8}\end{align}\) - step12: Solve the equation for \(x:\) \(\begin{align}&x=\log_{2}{\left(-1+\sqrt{1281}\right)}-3\\&x \notin \mathbb{R}\end{align}\) - step13: Find the union: \(x=\log_{2}{\left(-1+\sqrt{1281}\right)}-3\) The solutions to the given equations are: 1. \( x = 3 \) 2. \( x = 0 \) 3. \( x = 5 \) 4. \( x = 2 \) 5. \( x = \log_{2}(-1+\sqrt{1281})-3 \)

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Bonus Knowledge

To solve for \( x \) in each equation step-by-step: i. Start with \( 3^{x} - 3^{x-2} = 24 \). Rewrite it as \( 3^{x}(1 - \frac{1}{9}) = 24 \), so \( 3^{x} \cdot \frac{8}{9} = 24 \). Thus, \( 3^{x} = 24 \cdot \frac{9}{8} = 27 \). Therefore, \( x = 3 \) since \( 3^{3} = 27 \). ii. For \( 3^{x} + 3^{x+2} = 10 \), rewrite as \( 3^{x} + 9 \cdot 3^{x} = 10 \). Combine like terms: \( 10 \cdot 3^{x} = 10 \) gives \( 3^{x} = 1 \). Thus, \( x = 0 \). iii. The equation \( 2^{x} + 2^{x} + 2^{x} = 96 \) simplifies to \( 3 \cdot 2^{x} = 96 \). Dividing by 3: \( 2^{x} = 32 \) leads to \( x = 5 \) since \( 2^{5} = 32 \). iv. In \( 12^{x} + 12^{x} = 288 \), we rewrite it as \( 2 \cdot 12^{x} = 288 \). Dividing by 2 gives \( 12^{x} = 144 \), which means \( x = 2 \) since \( 12^{2} = 144 \). v. For \( 2^{x} + 4^{x+1} = 80 \), note that \( 4^{x+1} = 4 \cdot 4^{x} = 4 \cdot (2^{2})^{x} = 4 \cdot 2^{2x} \). This rewrites the equation to \( 2^{x} + 4 \cdot 2^{2x} = 80 \). Let \( 2^{x} = y \); then, the equation becomes \( y + 4y^{2} = 80 \). Rearranging gives \( 4y^{2} + y - 80 = 0 \). Using the quadratic formula results in two possible values for \( y \). This can be further simplified for finding \( x \). So the values of \( x \) are: i. \( x = 3 \) ii. \( x = 0 \) iii. \( x = 5 \) iv. \( x = 2 \) v. Solve that quadratic for values of \( x \). Happy solving!

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