Question

log0,8x+log0,8(x+1)=log0,8(x+3)

Answer 1

log0,8x+log0,8(x+1)=log0,8(x+3)

одз x + 3 > 0

x + 1 > 0

x > 0

x∈(0, +∞)

log0,8 x*(x+1)=log0,8 (x+3)

x^2 + x = x + 3

x² = 3

x12 = +- √3

-√3 < 0 не подходит

х = √3