Question

${8}^{x} + {8}^{ - x} = 9$

Answer 1

8ˣ + 1/8ˣ = 9

y + 1/y = 9

y² +1 = 9y

y² -9y +1 = 0

$\boxed{\mathsf{\Delta=b^2-4ac}}\\(-9)^{2} -4 * 1 * 1\\81 -4\\77\\\\\boxed{\mathsf{x=\frac{-b\pm\sqrt{\Delta}}{2a}}}}\\\frac{-(-9) \pm \sqrt{77} }{2} \\ \frac{9 \pm \sqrt{77} }{2}$

$8^{x} = \frac{9 \pm \sqrt{77} }{2} \to ln(8^{x}) = ln(\frac{9 \pm \sqrt{77} }{2}) \to x * ln(2^{3}) = ln(\frac{9 \pm \sqrt{77} }{2}) \to x = \frac{ln(\frac{9 \pm \sqrt{77} }{2})}{3 *ln(2)}$