Question

log in baza 2 (x+1) + log in baza 2 (x+7 )= 4

Answer 1

[tex]\log_2(x+1)(x+7)=4\Rightarrow (x+1)(x+7)=2^4 \\ \Rightarrow x^2+7x+8=16 \Rightarrow x^2+7x-8=0\Rightarrow x=-8,x=1[/tex]
se ia in considereare doar x=1 pentru care logaritmii sundt definiti.

Answer 2

   
$Folosim \;formula: \\ log_a b + log_a c = log_a(b*c) \;\;\; unde\;\;\; b\ \textgreater \ 0,\;c \ \textgreater \ 0 \\ \\ log_2(x+1)+log_2(x+7) =4 \\ log_2(x+1)(x+7)=4 \\ (x+1)(x+7)=2^4 \\ x^{2} +x+7x+7=16 \\ x^{2} +8x+7-16=0 \\ x^{2} +8x-9=0 \\ \\ x_{12}= \frac{-8 \pm \sqrt{64+4*9} }{2}= \frac{-8 \pm \sqrt{64+36} }{2}= \frac{-8 \pm \sqrt{100} }{2}=\frac{-8 \pm10 }{2}=-4 \pm 5 \\ \\ x_1 =-4 + 5 = \boxed{1} \\ x_2 = -4-5 = -9 \;\;\; \text{Aceasta solutie o vom elimina deoarece incalca } \\ \text{conditiile initiale ale formulei.}$