Question

Resolva em IR a equação: log2 x + log2 (x-2) - log2 (x-3) = 3
log x + log2 (x-2) - log2 (x-3) = 3 na base 2

Answer 1

Explicação passo-a-passo:

$\text{log}_{2}~x+\text{log}_{2}~(x-2)-\text{log}_{2}~(x-3)=3$

$\text{log}_{2}~\left[\dfrac{x\cdot(x-2)}{x-3}\right]=3$

$\dfrac{x\cdot(x-2)}{x-3}=2^3$

$\dfrac{x^2-2x}{x-3}=8$

$x^2-2x=8\cdot(x-3)$

$x^2-2x=8x-24$

$x^2-2x-8x+24=0$

$x^2-10x+24=0$

$\Delta=(-10)^2-4\cdot1\cdot24$

$\Delta=100-96$

$\Delta=4$

$x=\dfrac{-(-10)\pm\sqrt{4}}{2\cdot1}=\dfrac{10\pm2}{2}$

$x'=\dfrac{10+2}{2}~\longrightarrow~x'=\dfrac{12}{2}~\longrightarrow~x'=6$

$x"=\dfrac{10-2}{2}~\longrightarrow~x"=\dfrac{8}{2}~\longrightarrow~x"=4$

$S=\{4,6\}$