Question

9^x+3^x=272
Va rog să mă ajutați

Answer 1

${9}^{x} + {3}^{x} = 272$

${9}^{x} + {3}^{x} - 272 = 0$

${ ({3}^{x}) }^{2} +{3}^{x} - 272 = 0$

${3}^{x} = t \: \: ,t > 0$

${t}^{2} + t - 272 = 0$

$a = 1$

$b = 1$

$c = - 272$

$\Delta = {b}^{2} - 4ac$

$\Delta = {1}^{2}- 4 \times 1 \times ( - 272)$

$\Delta = 1 + 1088$

$\Delta = 1089$

$t_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}$

$t_{1,2}=\frac{- 1\pm\sqrt{1089}}{2 \times 1}$

$t_{1,2}=\frac{-1 \pm33}{2}$

$t_{1}=\frac{-1 + 33}{2} = \frac{32}{2} = 16> 0$

$= > {3}^{x} = 16$

$ln {3}^{x} = ln16$

$xln3 = ln16$

$x = \frac{ln16}{ln3}$

$t_{2}=\frac{-1 - 33}{2} = - \frac{34}{2} = - 17 < 0 = > \: nu \: verifica \: conditia$