Question

determinar o conjunto solução da equação logaritmica log3 (X+7)-log3 (X-1)=2

Answer 1

$log_3 (x+7)-log_3 (x-1)=2 \\ \\ \\ Log_3\left( \dfrac{x+7}{x-1}\right)=2 \\ \\ \\ Log_aB=c \to a^c= B \\ \\ \\ Log_3\left( \dfrac{x+7}{x-1}\right)=2 \to 3^2= \left( \dfrac{x+7}{x-1}\right) \\ \\ \\ Log_3\left( \dfrac{x+7}{x-1}\right)=2 \to 9= \left( \dfrac{x+7}{x-1}\right) \\ \\ \\ Log_3\left( \dfrac{x+7}{x-1}\right)=2 \to 9(x-1)=x+7 \\ \\ \\ Log_3\left( \dfrac{x+7}{x-1}\right)=2 \to 9x-9=x+7 \\ \\ \\ Log_3\left( \dfrac{x+7}{x-1}\right)=2 \to 9x-x=7+9$

$Log_3\left( \dfrac{x+7}{x-1}\right)=2 \to 8x=16 \\ \\ \\ Log_3\left( \dfrac{x+7}{x-1}\right)=2 \to x= \dfrac{16}{8} \\ \\ \\ \boxed{ \boxed{Log_3\left( \dfrac{x+7}{x-1}\right)=2 \to x=2 }}$

Answer 2

log3 (X+7)-log3 (X-1)=2

log   (X+7) = 2
     3 (X-1)

x + 7 = 3^2 ==> x + 7 = 9(x-1)
x - 1

x + 7 = 9x - 9

9x - x = 7 + 9

8x = 16

x = 2