Question

resolver as seguintes equações log2 (x²+2x-7)-log2(x-1)=2

Answer 1

log2 (x^2 + 2x - 7 / x - 1) = 2

x^2 + 2x - 7 / x - 1 = 2^2

x^2 + 2x - 7 = 4 (x - 1)

x^2 + 2x - 7 - 4x + 4 = 0

x^2 - 2x - 3 = 0

D = b^2 - 4ac

D = (-2)^2 - 4 . 1 . (-3)

D = 4 + 12 = 16

x = -b +- VD / 2a

x = -(-2) +-V16 / 2 . 1

x = 2 +- 4 / 2

x' = 2 + 4 / 2 = 6/2 = 3 ←←

x" = 2 - 4 / 2 = -2 / 2 = -1 ←←

obs.: nooossa... é verdade... precisa fazer a C.E. do logaritimando... putz, foi mal, siga a resposta abaixo, pfv! ;-)

Answer 2

Olá Dressamark,

Primeiro vamos checar a condição de existência dos logaritmandos:

$\mathsf{\ell og_2(x^2+2x-7)}$

$\mathsf{x^2+2x-7 \ \textgreater \ 0}\\\\\\\mathsf{\Delta=b^2-4ac}\\\mathsf{\Delta=2^2-4.1.(-7)}\\\mathsf{\Delta=4+28}\\\mathsf{\Delta=32}\\\\\\\mathsf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\Rightarrow x=\dfrac{-2\pm\sqrt{32}}{2.1}}\\\\\\\\\mathsf{x^+ \ \textgreater \ \dfrac{-2+4\sqrt{2}}{2}\Rightarrow x^+\ \textgreater \ \dfrac{\diagup\!\!\!\!2\cdot(-1+2\sqrt{2})}{\diagup\!\!\!\!2}\Rightarrow {\boxed{\mathsf{x^+\ \textgreater \ \!\!-1+2\sqrt{2} }}}\approx\boxed{\mathsf{1,8}}}$

$\mathsf{x^->\dfrac{-2-4\sqrt{2}}{2}\Rightarrow x^->\dfrac{\diagup\!\!\!\!2\cdot(-1-2\sqrt{2})}{\diagup\!\!\!\!2}\Rightarrow \boxed{\mathsf{x^->-1-2\sqrt{2}}}\approx\boxed{\mathsf{-3,8}}}}$

$\mathsf{\ell og_2(x-1)}\\\\\mathsf{x-1\neq0\Rightarrow \boxed{\mathsf{x> 1}}}$

Fazendo a interseção de x:

$\mathsf{x\ \textgreater \ }\begin{cases}\mathsf{-3,8~~~~~~\underline{---}\underset{-3,8}\circ\underline{+++}\underset{1}\circ\underline{+++}\underset{1,8}\circ\underline{+++}_\blacktriangleright}\\\\\mathsf{1~~~~~~~~~~~\underline{---}\underset{-3,8}\circ\underline{---}\underset{1}\circ\underline{+++}\underset{1,8}\circ\underline{+++}_\blacktriangleright}\\\\\mathsf{1,8~~~~~~~~\underline{---}\underset{-3,8}\circ\underline{---}\underset{1}\circ\underline{---}\underset{1,8}\circ\underline{+++}_\blacktriangleright}\end{cases}$

$\large\begin{array}\mathsf{\mathsf{C.E:\{}\mathsf{x\in\mathbb{R}~|~x\ \textgreater \ -1+2\sqrt{2}\}}}\end{array}$

Resolvendo a equação:

$\mathsf{\ell og_2 (x^2+2x-7)-\ell og_2(x-1)=2\Rightarrow \ell og_2\Big(\dfrac{x^2+2x-7}{x-1}\Big)=2}\\\\=\\\\\mathsf{\dfrac{x^2+2x-7}{x-1}=2^2\Rightarrow x^2+2x-7=4\cdot(x-1)\Rightarrow x^2+2x-7=4x-4}\\\\=\\\\\mathsf{x^2+2x-4x-7+4=0\Rightarrow \boxed{\mathsf{x^2-2x-3=0}}}\\\\\\\mathsf{\Delta=(-2)^2-4.1.(-3)}\\\mathsf{\Delta=4+12}\\\mathsf{\Delta=16}\\\\\\\mathsf{x^+=\dfrac{-(-2)+\sqrt{16}}{2.1}\Rightarrow x^+=\dfrac{2+4}{2}\Rightarrow x^+=\dfrac{6}{2}\Rightarrow \boxed{\mathsf{x^+=3}}}$

$\mathsf{x^-=\dfrac{-(-2)-\sqrt{16}}{2.1}\Rightarrow x^-=\dfrac{2-4}{2}\Rightarrow x^-=\dfrac{-2}{2}\Rightarrow \boxed{\mathsf{x^-=-1}}}$

$\large\boxed{\mathsf{S:x=3}}$

Dúvidas? comente