\( \left\{\begin{array}{l}3^{k x}, \text { for a suitable constant } k . \\ \text { 1. } 4^{x},(\sqrt{3})^{x},\left(\frac{1}{9}\right)^{x} \\ \text { (3. } 8^{2 x / 3}, 9^{3 x / 2}, 16^{-3 x / 4} \\ \text { (5. }\left(\frac{1}{4}\right)^{2 x},\left(\frac{1}{8}\right)^{-3 x},\left(\frac{1}{81}\right)^{x / 2}\end{array}\right. \)

It looks like you want to express each of the given exponential functions in the form \( 3^{k x} \) by finding the suitable constant \( k \). Let's analyze each expression one by one. ### 1. \( 4^{x}, \ (\sqrt{3})^{x}, \ \left(\frac{1}{9}\right)^{x} \) #### a. \( 4^{x} \) First, express 4 as a power of 3: \[ 4 = 3^{\log_3 4} \] So, \[ 4^{x} = \left(3^{\log_3 4}\right)^{x} = 3^{x \cdot \log_3 4} \] **Thus, \( k = \log_3 4 \).** #### b. \( (\sqrt{3})^{x} \) \[ \sqrt{3} = 3^{1/2} \] So, \[ (\sqrt{3})^{x} = \left(3^{1/2}\right)^{x} = 3^{(1/2) x} \] **Thus, \( k = \frac{1}{2} \).** #### c. \( \left(\frac{1}{9}\right)^{x} \) \[ \frac{1}{9} = 9^{-1} = (3^{2})^{-1} = 3^{-2} \] So, \[ \left(\frac{1}{9}\right)^{x} = \left(3^{-2}\right)^{x} = 3^{-2x} \] **Thus, \( k = -2 \).** ### 3. \( 8^{\frac{2x}{3}}, \ 9^{\frac{3x}{2}}, \ 16^{-\frac{3x}{4}} \) #### a. \( 8^{\frac{2x}{3}} \) \[ 8 = 2^{3} = \left(3^{\log_3 2}\right)^{3} = 3^{3 \log_3 2} \] So, \[ 8^{\frac{2x}{3}} = \left(3^{3 \log_3 2}\right)^{\frac{2x}{3}} = 3^{2x \cdot \log_3 2} \] **Thus, \( k = 2 \log_3 2 \).** #### b. \( 9^{\frac{3x}{2}} \) \[ 9 = 3^{2} \] So, \[ 9^{\frac{3x}{2}} = \left(3^{2}\right)^{\frac{3x}{2}} = 3^{3x} \] **Thus, \( k = 3 \).** #### c. \( 16^{-\frac{3x}{4}} \) \[ 16 = 2^{4} = \left(3^{\log_3 2}\right)^{4} = 3^{4 \log_3 2} \] So, \[ 16^{-\frac{3x}{4}} = \left(3^{4 \log_3 2}\right)^{-\frac{3x}{4}} = 3^{-3x \cdot \log_3 2} \] **Thus, \( k = -3 \log_3 2 \).** ### 5. \( \left(\frac{1}{4}\right)^{2x}, \ \left(\frac{1}{8}\right)^{-3x}, \ \left(\frac{1}{81}\right)^{\frac{x}{2}} \) #### a. \( \left(\frac{1}{4}\right)^{2x} \) \[ \frac{1}{4} = 4^{-1} = \left(3^{\log_3 4}\right)^{-1} = 3^{-\log_3 4} \] So, \[ \left(\frac{1}{4}\right)^{2x} = \left(3^{-\log_3 4}\right)^{2x} = 3^{-2x \cdot \log_3 4} \] **Thus, \( k = -2 \log_3 4 \).** #### b. \( \left(\frac{1}{8}\right)^{-3x} \) \[ \frac{1}{8} = 8^{-1} = \left(3^{\log_3 8}\right)^{-1} = 3^{-\log_3 8} \] So, \[ \left(\frac{1}{8}\right)^{-3x} = \left(3^{-\log_3 8}\right)^{-3x} = 3^{3x \cdot \log_3 8} \] **Thus, \( k = 3 \log_3 8 \).** #### c. \( \left(\frac{1}{81}\right)^{\frac{x}{2}} \) \[ \frac{1}{81} = 81^{-1} = \left(3^{4}\right)^{-1} = 3^{-4} \] So, \[ \left(\frac{1}{81}\right)^{\frac{x}{2}} = \left(3^{-4}\right)^{\frac{x}{2}} = 3^{-2x} \] **Thus, \( k = -2 \).** --- ### Summary of Constants \( k \) 1. **First Set:** - \( 4^{x} = 3^{(\log_3 4) x} \) → \( k = \log_3 4 \) - \( (\sqrt{3})^{x} = 3^{\frac{1}{2} x} \) → \( k = \frac{1}{2} \) - \( \left(\frac{1}{9}\right)^{x} = 3^{-2x} \) → \( k = -2 \) 3. **Third Set:** - \( 8^{\frac{2x}{3}} = 3^{2 (\log_3 2) x} \) → \( k = 2 \log_3 2 \) - \( 9^{\frac{3x}{2}} = 3^{3x} \) → \( k = 3 \) - \( 16^{-\frac{3x}{4}} = 3^{-3 (\log_3 2) x} \) → \( k = -3 \log_3 2 \) 5. **Fifth Set:** - \( \left(\frac{1}{4}\right)^{2x} = 3^{-2 (\log_3 4) x} \) → \( k = -2 \log_3 4 \) - \( \left(\frac{1}{8}\right)^{-3x} = 3^{3 (\log_3 8) x} \) → \( k = 3 \log_3 8 \) - \( \left(\frac{1}{81}\right)^{\frac{x}{2}} = 3^{-2x} \) → \( k = -2 \) --- **Note:** \( \log_3 a \) represents the logarithm of \( a \) with base 3. Depending on the context or the specific requirements of your problem, you might need to simplify these logarithmic expressions further or leave them as is.