Question

Determine o valor real de x que torna verdadeira a
igualdade 3^(2x) – 10 · 3^x + 9 = 0

Answer 1

$3 x^{2x-10} \cdot 3 x^{x+9} = 0$
$2x - 10 + x + 9 = 0$
$x = \frac{1}{3}$

Answer 2

$3^{2x}-10.3^{x}+9 = 0 \\ \\ (3^{x})^{2}-10.3^{x}+9 = 0 \\ \\ 3^{x} = y \\ \\ y^{2}-10y+9 = 0 \\ \\ delta = (-10)^{2} - 4 . 1 . 9 = 100 - 36 = 64 \\ \\ y = \frac{-(-10)+- \sqrt{64} }{2.1} \\ \\ y = \frac{10+- 8 }{2} \\ \\ y_{1}= \frac{10+ 8 }{2} = \frac{18 }{2} = 9 \\ \\ y_{2}= \frac{10- 8 }{2} = \frac{2 }{2} = 1$

$y = 3^{x} \\ 9 = 3^{x} \\ 3^{2} = 3^{x} \\ \\ x = 2 \\ \\ 1 = 3^{x} \\ 3^{0} = 3^{x} \\ \\ x = 0$

Logo, x = 0 e x = 2