4.6. \( 3 \times 3^{x}+9 \times 3^{-x}-28=0 \)

To solve the equation \( 3 \times 3^{x}+9 \times 3^{-x}-28=0 \), we can first rewrite it in terms of \( 3^x \). Notice that \( 9 = 3^2 \), and thus \( 9 \times 3^{-x} = 3^2 \times 3^{-x} = 3^{2-x} \). We can let \( y = 3^x \), which transforms the equation to \( 3y + \frac{9}{y} - 28 = 0 \) by rewriting \( 3^{-x} \) as \( \frac{1}{y} \). Multiplying through by \( y \) to eliminate the fraction gives \( 3y^2 - 28y + 9 = 0 \). This is a quadratic equation that can be solved using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Calculating the discriminant: \( D = (-28)^2 - 4 \cdot 3 \cdot 9 = 784 - 108 = 676 \). Now, applying the formula: \[ y = \frac{28 \pm \sqrt{676}}{6} = \frac{28 \pm 26}{6}. \] This gives two possible solutions for \( y \): 1. \( y = \frac{54}{6} = 9 \) 2. \( y = \frac{2}{6} = \frac{1}{3} \) Recalling that \( y = 3^x \), we have: 1. \( 3^x = 9 \) implies \( x = 2 \) 2. \( 3^x = \frac{1}{3} \) implies \( x = -1 \) Thus, the solutions to the equation are \( x = 2 \) and \( x = -1 \).