Question

the roots of equation 3^x+2+3^-x=10 are​

Answer 1

Answer:

${3}^{x} + 2 + \frac{1}{ {3}^{x} } = 10 \\ let \: {3}^{x} = y \\ y + 2 + \frac{1}{y} = 10 \\ {y}^{2} + 2y + 1 = 10y \\ {y}^{2} - 8y + 1 = 0 \\ y = \frac{8 + \sqrt{ {8}^{2} - 4} }{2} \: or \: \frac{8 - \sqrt{ {8}^{2} - 4 } }{2} \\ y = \frac{8 + 2 \sqrt{15} }{2} \: or \: \frac{8 - 2 \sqrt{15} }{2} \\ y = 4 + \sqrt{15} \: or \: 4 - \sqrt{15} \\ i.e \: \: {3}^{x} = 4 + \sqrt{15} \: or \: 4 - \sqrt{15} \\ x = log_{3}(4 + \sqrt{15} ) \: \: or \: \: log_{3}(4 - \sqrt{15} ) \\ ans.$