Question

log1/2 (x^2 - 1) - log1/2 (x - 2) = -3
log (x^2 - 1) - log1/2 (x - 2) = -3 na base 1/2 ,ajudem por favor

Answer 1

Explicação passo-a-passo:

$\text{log}_{\frac{1}{2}}~(x^2-1)-\text{log}_{\frac{1}{2}}~(x-2)=-3$

$\text{log}_{\frac{1}{2}}~\left(\dfrac{x^2-1}{x-2}\right)=-3$

$\dfrac{x^2-1}{x-2}=\left(\dfrac{1}{2}\right)^{-3}$

$\dfrac{x^2-1}{x-2}=8$

$x^2-1=8\cdot(x-2)$

$x^2-1=8x-16$

$x^2-8x-1+16=0$

$x^2-8x+15=0$

$\Delta=(-8)^2-4\cdot1\cdot15$

$\Delta=64-60$

$\Delta=4$

$x=\dfrac{-(-8)\pm\sqrt{4}}{2\cdot1}=\dfrac{8\pm2}{2}$

$x'=\dfrac{8+2}{2}~\longrightarrow~x'=\dfrac{10}{2}~\longrightarrow~x'=5$

$x"=\dfrac{8-2}{2}~\longrightarrow~x"=\dfrac{6}{2}~\longrightarrow~x"=3$

$S=\{3,5\}$