Question

Resolva a equação log2 (x – 1) + 1 = log2 (x + 2) + log2 (7 – x) – log2 (3)
Resposta = 4

Answer 1

Olá Leonardo,


Antes de começar a resolver a equação precisamos verificar a condição de existência dos logaritmandos


$\mathsf{C.E\begin{cases}x-1\ \textgreater \ 0~|~x\ \textgreater \ 1\\x+2\ \textgreater \ 0~|~x\ \textgreater \ -2\\7-x\ \textgreater \ 0~|~x\ \textless \ 7\end{cases}}\\\\\\\mathsf{S=\{x\in\mathbb{R}~|~1\ \textless \ x\ \textless \ 7\}}\\\\\\\log_2 (x-1)+1=\log_2 (x+2)+\log_2 (7-x)-\log_2(3)\\\log_2(x-1)+\log_2(2)=\log_2\big(\frac{(x+2).(7-x)}{3}\big)\\\log_2(2x-2)=\log_2(\frac{7x-x^2+14-2x}{3})\\\\2x-2=\frac{7x-x^2+14-2x}{3}~*(3)\\\\6x-6=-x^2+5x+14\\x^2+x-20=0$



$\Delta=b^2 - 4.a.c\\\Delta=1^2-4.1.(-20)\\\Delta=81\\\\\\x=\frac{-(1)\pm\sqrt{81}}{2.(1)}\\\\x=\frac{-1+9}{2}\\\\x=\frac{8}{2}\\\\\boxed{x=4}\gets~\mathsf{Atende~a~condic\~ao~de~exist\^encia}\\\\\\x'=\frac{-1-9}{-2}\\\\x'=\frac{-10}{2}\\\\\\\boxed{x'=-5}\gets~\mathsf{N\~ao~atende~a~condic\~ao~de~exist\^encia}\\\\\\\\\Large\boxed{S=4}$


Dúvidas? comente

Answer 2

Temos que as bases são as mesmas e que 1= Log2(2) logo 

(x-1).2=(x+2).(7-x)/3
2x-2=7x-x²+14-2x/3
6x-6=7x-x²+14-2x
6x-7x+2x=14+6-x²
x²+x-20=0

Δ=1²-4.1(-20)
Δ=1+80
Δ=81

x=-1+-√81/2
x=-1+-9/2
X'=-1+9/2= 8/2=  4 
x''=-1-9/2 = -10/2 = -5

Condição de existência será solução: 4