Question

log in baza 2 din x + log in baza 2 din x-1 = log in baza 2din 12
Mersi mult !

Answer 1

$log_{2}(x) + log_{2}(x - 1) = log_{2}12$
$x \times (x - 1) = 12$
x²-x=12
x²-x-12=0
∆=(-1)²-4•1•-12=1+48=49
x1=(1+7)/2=8/2=4
x2=(1-7)/2=-6/2=-3
S={-3,4}

Answer 2

log₂x + log₂(x-1) = log₂ 12 <=> log₂(x)(x-1) = log₂12 =>

x(x-1) = 12 <=> x²-x-12=0 => x₁=[1+√(1+48)]/2 = (1+7)/2 = 4

x₂ = (1-7)/2 = -3 ; din conditia logaritmului => x>0 =>x₂ nu este solutie valabila

=> Solutie unica : x= 4