Question

Хееелп!:) log2(3^x-1)+log2(3^x-2)=1

Answer 1

3^x -1 =a
3^x-2 = b 
log2 a + log2b = 1
log2 ab=1
(3^x-1)(3^x-2)=2
3^x = t
(t-1)(t-2) =2
t^2 -3t+1=0
D=(-3)^2 - 4 *1*1=9 - 4 = 5
√D=√5
t 1= 3+√5 / 2 
t 2 = 3 -√5 / 2 

Answer 2

Спасибо большое)