Question

log1,5 (x-0,5)+log1,5(x+0,25=log1,5 (x2-1,75)+1

Answer 1

Log1,5(x-0,5) + log1,5(x+0,25) = log1,5(x² - 1,75) + 1
Log1,5(x-0,5) + log1,5(x+0,25) = log1,5(x² - 1,75) + log1,5 (1,5)
log1,5[(x-0,5).(x+0,25)] = log1,5[(x²-1,75).1,5]
log1,5[x² + 0,25x - 0,5x - 0,125] - log1,5[1,5x² - 2,625] = 0
log1,5[x² - 0,25x - 0,125] - log1,5[1,5x² - 2,625] = 0
log1,5[(x² - 0,25x - 0,125)/(1,5x² - 2,625)] = 0

Logₐn = 0, então n = 1

(x² - 0,25x - 0,125)/(1,5x² - 2,625) = 1
x² - 0,25x - 0,125 = 1,5x² - 2,625
- x² + 0,25x + 0,125 + 1,5x² - 2,625 = 0
0,5x² + 0,25x - 2,5 = 0
x² + 0,5x - 5 = 0

Δ = b² - 4ac
Δ = 0,5² + 20
Δ = 20,25

x = (-b +-Δ)/2a
x = (-0,5 +- 4,5)/2

x' = (-0,5 + 4,5) / 2 = 4/2 = 2
x" = (-0,5 - 4,5) / 2 = -5/2 = -2,5

Log1,5(x-0,5) + log1,5(x+0,25) = log1,5(x² - 1,75) + 1
Log1,5(2-0,5) + log1,5(2+0,25) = log1,5(2² - 1,75) + 1
Log1,5(1,5) + log1,5(2,25) = log1,5(2,25) + 1
1 + 2 = 2 + 1

Log1,5[(-2,5) - 0,5] + log1,5[(-2,5) + 0,25] = log1,5[(-2,5²) - 1,75] + 1
Log1,5(-3) + log1,5(-2,25) = log1,5(4,5) + 1

Logₓ n => x≠1 e n>0

Portanto

S = {x = 2}

=)