Question

Log₂(9 x-1 +7) = 2 + log₂(3 x-1 +1)

tylko ze to x-1 jest w potędze

Answer 1

Log₂(9 x-1 +7)= 2 + log₂(3 x-1 +1)
log₂(9 x-1 +7)= log₂4 + log₂(3 x-1 +1)
log₂(9 x-1 +7)= log₂ 4(3 x-1 +1)
9 x-1 +7 = 4(3 x-1 +1)
9 x-1 +7 = 4* 3 x-1 + 4
9 x-1 - 4* 3 x-1 + 3=0

3 x-1 = m
m² - 4m +3=0
Δ=16-12=4
√Δ=2
m₁=4-2/2= 1
m₂ = 4+ 2/2= 3

3 (x-1) = 3 V 3 (x-1) = 1
x-1 = 1 V x-1=0
x=2 V x=1