Question

Resolva em IR a equação logarítmica

$2log_2(x+5)+log_{0,5}(-x+7)=4$

Answer 1

$2log_{2}(x+5)+log_{0,5}(-x+7)=4\\log_{2}(x+5)^{2}+log_{(2^{-1})}(-x+7)=4\\log_{2}(x+5)^{2}+(\frac{1}{-1}).log_{2}(-x+7)=4\\log_{2}(x+5)^{2}-log_{2}(-x+7)=4$

Aplicando a propriedade:

$log_{2}\left[\dfrac{(x+5)^{2}}{-x+7}\right]=4\\\\\\\dfrac{(x+5)^{2}}{-x+7}=2^{4}\\\\\\\dfrac{x^{2}+10x+25}{-x+7}=16\\\\\\x^{2}+10x+25=16(-x+7)\\\\x^{2}+10x+25=-16x+112\\\\x^{2}+10x+16x+25-112=0\\\\x^{2}+26x-87=0$
____

$\Delta=b^{2}-4ac\\\Delta=26^{2}-4.1.(-87)\\\Delta=676+348\\\Delta=1024$

$x=\dfrac{-b\pm\sqrt{\Delta}}{2a}~~~\therefore~~~\dfrac{-26\pm\sqrt{1024}}{2.1}~~~\therefore~~~\boxed{x=-13\pm16}$

Logo:

$x'=-13+16=3\\x''=-13-16=-29$

Descartamos x = -29, pois teríamos um logaritmando negativo na expressão

$\boxed{\boxed{S=\{3\}}}$

Answer 2

Resposta:

$x = 3$

Explicação passo-a-passo:

$2 log_{2}(x + 5) + log_{0,5}( - x + 7) = 4 \\ 2 log_{2}(x + 5) + log_{0,5}( - x + 7) = 4, \: e \: ( - 5,7) \\ log_{2}((x + 5 {)}^{2} ) + log_{ {2}^{ - 1} }( - x + 7) = 4 \\ log_{2}((x + 5 {)}^{2} ) - log_{2}( - x + 7) = 4 \\ log_{2}( \frac{(x + 5 {)}^{2} }{ - x + 7} ) = 4 \\ \frac{(x + 5{)}^{2} }{ - x + 7} = {2}^{4} \\ \frac{ {x}^{2} + 10x + 25 }{ - x + 7} = 16 \\ {x}^{2} + 10x + 25 = 16( - x + 7) \\ {x}^{2} + 10x + 25 - 16( - x + 7) = 0 \\ {x}^{2} + 10x + 25 + 16x - 112 = 0 \\ {x}^{2} + 26x - 87 = 0 \\ x = \frac{ - 26 ± \sqrt{ {26}^{2} - 4 \times 1 \times ( - 87) } }{2 \times 1} \\ x = \frac{ - 26± \sqrt{676 + 348} }{2} \\ x = \frac{ - 26± \sqrt{1024} }{2} \\ x = \frac{ - 26± 32 }{2} \\ \\ x = \frac{ - 26 + 32}{2} \\ x = \frac{ - 26 - 32}{2} \\ \\ x = 3 \\ x = - 29 , \: x \: e \: ( - 5,7) \\ \\ x = 3$

• Espero ter ajudado.