Question

Alguém sabe???
log x+1 19-x=2

Answer 1

$log_{x+1}(19-x) = 2$

$(x+1)^2 = 19-x$

$x^2+2x+1 = 19-x$

$x^2+2x+1-19+x = 0$

$x^2 + 3x-18 = 0$

$\Delta = (3)^2-4(1)(-18)$

$\Delta = 9 + 72$

$\Delta = 81$

$\sqrt{\Delta} = 9$

$x = \frac{-3 +- 9}{2}$

$\boxed{x' = -6}$

$\boxed{x'' = 3}$

$S=(-6,3)$

Espero ter ajudado. :))