EXERCISE 7 Solve for \( x \) : \( \begin{array}{lll}\text { (a) } 3^{x}=1 & \text { (b) } 2^{x}=\sqrt{2} & \text { (c) } 4^{x}=8 \\ \text { (d) }\left(\frac{1}{9}\right)^{x}=27 & \text { (e) } 16 \cdot 16^{x}=64 & \text { (f) } 3^{x+1}=\frac{\sqrt{n}}{3} \\ \text { (g) } 4\left(\frac{1}{4}\right)^{x-1}=8 & \text { (h) } \frac{1}{3}(3)^{x-1}=\frac{1}{27} & \text { (i) } 8^{x} \cdot 16^{x-1} \\ \text { (i) } \frac{3^{2 x-1}}{3^{x}}=3 & \text { (k) } \sqrt[3]{9}=(\sqrt{3})^{2 x} & \text { (l) } \\ \text { (m) } 8^{-x}-2.4^{x-1}=0 & \text { (n) } 6^{x}+6^{x}+6^{x}+6^{x}+6^{x}+6^{x}=6^{6 x}\end{array} \)

Solve the equation by following steps: - step0: Solve for \(x\): \(3^{x}=1\) - step1: Rewrite in exponential form: \(3^{x}=3^{0}\) - step2: Set the exponents equal: \(x=0\) Solve the equation \( 2^{x}=\frac{1}{2} \). Solve the equation by following steps: - step0: Solve for \(x\): \(2^{x}=\frac{1}{2}\) - step1: Rewrite the expression: \(2^{x}=2^{-1}\) - step2: Set the exponents equal: \(x=-1\) Solve the equation \( 4^{x}=8 \). Solve the equation by following steps: - step0: Solve for \(x\): \(4^{x}=8\) - step1: Rewrite in exponential form: \(2^{2x}=2^{3}\) - step2: Set the exponents equal: \(2x=3\) - step3: Divide both sides: \(\frac{2x}{2}=\frac{3}{2}\) - step4: Divide the numbers: \(x=\frac{3}{2}\) Solve the equation \( \frac{1}{9^{x}}=27 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{1}{9^{x}}=27\) - step1: Rearrange the terms: \(9^{-x}=27\) - step2: Rewrite in exponential form: \(3^{-2x}=3^{3}\) - step3: Set the exponents equal: \(-2x=3\) - step4: Change the signs: \(2x=-3\) - step5: Divide both sides: \(\frac{2x}{2}=\frac{-3}{2}\) - step6: Divide the numbers: \(x=-\frac{3}{2}\) Solve the equation \( \frac{3^{2x-1}}{3^{x}}=3 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{3^{2x-1}}{3^{x}}=3\) - step1: Simplify: \(3^{x-1}=3\) - step2: Set the exponents equal: \(x-1=1\) - step3: Move the constant to the right side: \(x=1+1\) - step4: Add the numbers: \(x=2\) Solve the equation \( \frac{1}{3} \times 3^{x-1}=\frac{1}{27} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{1}{3}\times 3^{x-1}=\frac{1}{27}\) - step1: Multiply the terms: \(3^{x-2}=\frac{1}{27}\) - step2: Rewrite in exponential form: \(3^{x-2}=3^{-3}\) - step3: Set the exponents equal: \(x-2=-3\) - step4: Move the constant to the right side: \(x=-3+2\) - step5: Add the numbers: \(x=-1\) Solve the equation \( 16 \times 16^{x}=64 \). Solve the equation by following steps: - step0: Solve for \(x\): \(16\times 16^{x}=64\) - step1: Multiply the terms: \(16^{x+1}=64\) - step2: Rewrite in exponential form: \(2^{4\left(x+1\right)}=2^{6}\) - step3: Set the exponents equal: \(4\left(x+1\right)=6\) - step4: Divide both sides: \(\frac{4\left(x+1\right)}{4}=\frac{6}{4}\) - 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: \(x=\frac{1}{2}\) Solve the equation \( 8^{x} \times 16^{x-1}=1 \). Solve the equation by following steps: - step0: Solve for \(x\): \(8^{x}\times 16^{x-1}=1\) - step1: Multiply the terms: \(2^{7x-4}=1\) - step2: Rewrite in exponential form: \(2^{7x-4}=2^{0}\) - step3: Set the exponents equal: \(7x-4=0\) - step4: Move the constant to the right side: \(7x=0+4\) - step5: Remove 0: \(7x=4\) - step6: Divide both sides: \(\frac{7x}{7}=\frac{4}{7}\) - step7: Divide the numbers: \(x=\frac{4}{7}\) Solve the equation \( 6^{x}+6^{x}+6^{x}+6^{x}+6^{x}+6^{x}=6^{6x} \). Solve the equation by following steps: - step0: Solve for \(x\): \(6^{x}+6^{x}+6^{x}+6^{x}+6^{x}+6^{x}=6^{6x}\) - step1: Simplify: \(6^{x+1}=6^{6x}\) - step2: Set the exponents equal: \(x+1=6x\) - step3: Move the variable to the left side: \(x+1-6x=0\) - step4: Subtract the terms: \(-5x+1=0\) - step5: Move the constant to the right side: \(-5x=0-1\) - step6: Remove 0: \(-5x=-1\) - step7: Change the signs: \(5x=1\) - step8: Divide both sides: \(\frac{5x}{5}=\frac{1}{5}\) - step9: Divide the numbers: \(x=\frac{1}{5}\) Solve the equation \( 4 \times \frac{1}{4^{x-1}}=8 \). Solve the equation by following steps: - step0: Solve for \(x\): \(4\times \frac{1}{4^{x-1}}=8\) - step1: Simplify: \(4^{-x+2}=8\) - step2: Rewrite in exponential form: \(2^{2\left(-x+2\right)}=2^{3}\) - step3: Set the exponents equal: \(2\left(-x+2\right)=3\) - step4: Divide both sides: \(\frac{2\left(-x+2\right)}{2}=\frac{3}{2}\) - step5: Divide the numbers: \(-x+2=\frac{3}{2}\) - step6: Move the constant to the right side: \(-x=\frac{3}{2}-2\) - step7: Subtract the numbers: \(-x=-\frac{1}{2}\) - step8: Change the signs: \(x=\frac{1}{2}\) Solve the equation \( 8^{-x}-2 \times 4^{x-1}=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(8^{-x}-2\times 4^{x-1}=0\) - step1: Factor the expression: \(2^{-x}\left(1-2^{5x-1}\right)\times 4^{-x}=0\) - step2: Separate into possible cases: \(\begin{align}&2^{-x}=0\\&1-2^{5x-1}=0\\&4^{-x}=0\end{align}\) - step3: The statement is false for any value of \(x:\) \(\begin{align}&x \notin \mathbb{R}\\&x=\frac{1}{5}\\&x \notin \mathbb{R}\end{align}\) - step4: Find the union: \(x=\frac{1}{5}\) Here are the solutions for each part of the exercise: ### (a) \( 3^{x} = 1 \) To solve for \( x \): \[ x = 0 \] ### (b) \( 2^{x} = \sqrt{2} \) Rewriting \( \sqrt{2} \) as \( 2^{1/2} \): \[ x = -1 \] ### (c) \( 4^{x} = 8 \) Rewriting \( 8 \) as \( 2^{3} \) and \( 4 \) as \( 2^{2} \): \[ (2^{2})^{x} = 2^{3} \implies 2^{2x} = 2^{3} \implies 2x = 3 \implies x = \frac{3}{2} \] ### (d) \( \left(\frac{1}{9}\right)^{x} = 27 \) Rewriting \( \frac{1}{9} \) as \( 9^{-1} \) and \( 27 \) as \( 3^{3} \): \[ (3^{-2})^{x} = 3^{3} \implies 3^{-2x} = 3^{3} \implies -2x = 3 \implies x = -\frac{3}{2} \] ### (e) \( 16 \cdot 16^{x} = 64 \) Rewriting \( 64 \) as \( 16^{\frac{3}{2}} \): \[ 16^{1} \cdot 16^{x} = 16^{\frac{3}{2}} \implies 16^{x+1} = 16^{\frac{3}{2}} \implies x + 1 = \frac{3}{2} \implies x = \frac{1}{2} \] ### (f) \( 3^{x+1} = \frac{\sqrt{n}}{3} \) This equation cannot be solved without knowing the value of \( n \). ### (g) \( 4\left(\frac{1}{4}\right)^{x-1} = 8 \) Rewriting \( 8 \) as \( 4^{\frac{3}{2}} \): \[ 4 \cdot 4^{-(x-1)} = 4^{\frac{3}{2}} \implies 4^{1-(x-1)} = 4^{\frac{3}{2}} \implies 2 - x = \frac{3}{2} \implies x = -\frac{1}{2} \] ### (h) \( \frac{1}{3}(3)^{x-1} = \frac{1}{27} \) Rewriting \( \frac{1}{27} \) as \( 3^{-3} \): \[ \frac{1}{3} \cdot 3^{x-1} = 3^{-3} \implies 3^{x-1-1} = 3^{-3} \implies x - 2 = -3 \implies x = -1 \] ### (i) \( 8^{x} \cdot 16^{x-1} = 1 \) Rewriting \( 8 \) as \( 2^{3} \) and \( 16 \) as \( 2^{4} \): \[ (2^{3})^{x} \cdot (2^{4})^{x-1} = 1 \implies 2^{3x + 4(x-1)} = 1 \implies 3x + 4x - 4 = 0 \implies 7x = 4 \implies x = \frac{4}{7} \] ### (j) \( \frac{3^{2x-1}}{3^{x}} = 3 \) This simplifies to: \[ 3^{2x-1-x} = 3^{1} \implies 3^{x-1} = 3^{1} \implies x - 1 = 1 \implies x = 2 \] ### (k) \( \sqrt[3]{9} = (\sqrt{3})^{2x} \) Rewriting \( \sqrt[3]{9} \) as \( 3^{2/3} \): \[ 3^{2/3} = 3^{x} \implies \frac{2}{3} = 2x \implies x = \frac{1}{3} \] ### (l) \( 8^{-x} - 2 \cdot 4^{x-1} = 0 \) Rewriting \( 8 \) and \( 4 \): \[ (2^{3})^{-x} - 2 \cdot (2^{2})^{x-1} = 0 \implies 2^{-3x} - 2 \cdot 2^{2x-2} = 0 \implies 2^{-3x} - 2^{2x-1} = 0 \] This leads to: \[ -3x = 2x - 1 \implies -5x = -1 \implies x = \frac{1}{5} \] ### (m) \( 6^{x} + 6^{x} + 6^{x} + 6^{x} + 6^{x} + 6^{x} = 6^{6x} \) This simplifies to: \[ 6 \cdot 6^{x} = 6^{6x} \implies 6^{x+1} = 6^{6x} \implies x + 1 = 6x \implies 5x = 1 \implies x = \frac{1}{5} \] ### Summary of Solutions: - (a) \( x = 0 \) - (b) \( x = -1 \) - (c) \( x = \frac{3}{2} \) - (d) \( x = -\frac{3}{2} \) - (e) \( x = \frac{1}{2} \) - (f) Cannot be solved without \( n \) - (g) \( x = -\frac{1}{2} \) - (h) \( x = -1 \) - (i) \( x = \frac{4}{7} \) - (j) \( x = 2 \) - (k) \( x = \frac{1}{3} \) - (l) \( x = \frac{1}{5} \) - (m) \( x = \frac{1}{5} \)