Write each expression in Exercises \( 1-14 \) in the form \( 2^{k x} \) or \( 3^{k x} \), for a suitable constant \( k \). \( \begin{array}{ll}\text { 1. } 4^{x},(\sqrt{3})^{x},\left(\frac{1}{9}\right)^{x} & \text { 2. } 27^{x},(\sqrt[3]{2})^{x},\left(\frac{1}{8}\right)^{x} \\ \begin{array}{ll}\text { (3) } 8^{2 x / 3}, 9^{3 x / 2}, 16^{-3 x / 4} & \text { 4. } 9^{-x / 2}, 8^{4 x / 3}, 27^{-2 x / 3} \\ \text { 5. }\left(\frac{1}{4}\right)^{2 x},\left(\frac{1}{8}\right)^{-3 x},\left(\frac{1}{81}\right)^{x / 2} & \text { 6. }\left(\frac{1}{9}\right)^{2 x},\left(\frac{1}{27}\right)^{x / 3},\left(\frac{1}{16}\right)^{-x / 2}\end{array}\end{array} \)

Sure! Let's rewrite each expression in the form \( 2^{k x} \) or \( 3^{k x} \), where \( k \) is a suitable constant. ### **1.** a) \( 4^{x} \) Since \( 4 = 2^2 \): \[ 4^{x} = (2^2)^{x} = 2^{2x} \] b) \( (\sqrt{3})^{x} \) Since \( \sqrt{3} = 3^{1/2} \): \[ (\sqrt{3})^{x} = (3^{1/2})^{x} = 3^{\frac{x}{2}} \] c) \( \left(\frac{1}{9}\right)^{x} \) Since \( \frac{1}{9} = 3^{-2} \): \[ \left(\frac{1}{9}\right)^{x} = 3^{-2x} \] --- ### **2.** a) \( 27^{x} \) Since \( 27 = 3^3 \): \[ 27^{x} = (3^3)^{x} = 3^{3x} \] b) \( \left(\sqrt[3]{2}\right)^{x} \) Since \( \sqrt[3]{2} = 2^{1/3} \): \[ \left(\sqrt[3]{2}\right)^{x} = 2^{\frac{x}{3}} \] c) \( \left(\frac{1}{8}\right)^{x} \) Since \( \frac{1}{8} = 2^{-3} \): \[ \left(\frac{1}{8}\right)^{x} = 2^{-3x} \] --- ### **3.** a) \( 8^{\frac{2x}{3}} \) Since \( 8 = 2^3 \): \[ 8^{\frac{2x}{3}} = (2^3)^{\frac{2x}{3}} = 2^{2x} \] b) \( 9^{\frac{3x}{2}} \) Since \( 9 = 3^2 \): \[ 9^{\frac{3x}{2}} = (3^2)^{\frac{3x}{2}} = 3^{3x} \] c) \( 16^{-\frac{3x}{4}} \) Since \( 16 = 2^4 \): \[ 16^{-\frac{3x}{4}} = (2^4)^{-\frac{3x}{4}} = 2^{-3x} \] --- ### **4.** a) \( 9^{-\frac{x}{2}} \) Since \( 9 = 3^2 \): \[ 9^{-\frac{x}{2}} = (3^2)^{-\frac{x}{2}} = 3^{-x} \] b) \( 8^{\frac{4x}{3}} \) Since \( 8 = 2^3 \): \[ 8^{\frac{4x}{3}} = (2^3)^{\frac{4x}{3}} = 2^{4x} \] c) \( 27^{-\frac{2x}{3}} \) Since \( 27 = 3^3 \): \[ 27^{-\frac{2x}{3}} = (3^3)^{-\frac{2x}{3}} = 3^{-2x} \] --- ### **5.** a) \( \left(\frac{1}{4}\right)^{2x} \) Since \( \frac{1}{4} = 2^{-2} \): \[ \left(\frac{1}{4}\right)^{2x} = 2^{-4x} \] b) \( \left(\frac{1}{8}\right)^{-3x} \) Since \( \frac{1}{8} = 2^{-3} \): \[ \left(\frac{1}{8}\right)^{-3x} = 2^{9x} \] c) \( \left(\frac{1}{81}\right)^{\frac{x}{2}} \) Since \( \frac{1}{81} = 3^{-4} \): \[ \left(\frac{1}{81}\right)^{\frac{x}{2}} = 3^{-2x} \] --- ### **6.** a) \( \left(\frac{1}{9}\right)^{2x} \) Since \( \frac{1}{9} = 3^{-2} \): \[ \left(\frac{1}{9}\right)^{2x} = 3^{-4x} \] b) \( \left(\frac{1}{27}\right)^{\frac{x}{3}} \) Since \( \frac{1}{27} = 3^{-3} \): \[ \left(\frac{1}{27}\right)^{\frac{x}{3}} = 3^{-x} \] c) \( \left(\frac{1}{16}\right)^{-\frac{x}{2}} \) Since \( \frac{1}{16} = 2^{-4} \): \[ \left(\frac{1}{16}\right)^{-\frac{x}{2}} = 2^{2x} \] --- Each expression is now expressed as either \( 2^{k x} \) or \( 3^{k x} \), where \( k \) is a constant.