Question

Calculați log in baza 2 din (x^2-2x)=3

Answer 1

Răspuns:

Explicație pas cu pas:

$\texttt{Conditii de existenta:}x\in(-\infty,0)\cup(2,\infty)\\\log_2(x^2-2x)=3\\x^2-2x=2^3\\x^2-2x=8\\x^2-2x-8=0\\\Delta=4+4\cdot 8=36\Rightarrow \sqrt{\Delta}=6\\x_1=\dfrac{2+6}{2}=\dfrac{8}{2}=4\\\\x_2=\dfrac{2-6}{2}=-\dfrac{4}{2}=-2\\\texttt{Observam ca ambele solutii convin, prin urmare solutia este}\\S=\{-2,4\}$

Answer 2

log₂x²-2x=3

C.E.:x²-2x>0<=>x(x-2)>0

x        | -∞             0           2         +∞

x(x-2)|+++++++++0----------0+++++++

x∈(-∞;0)∪(2;+∞)

log₂x²-2x=3<=>x²-2x=2³<=>x²-2x-8=0

Δ=4+32=36

x₁₂=(2±6)/2

x₁=(2-6)/2=-4/2=-2

x₂=(2+6)/2=8/2=4

S={-2,4}