Question

Log 3(x+1)=2
Am teza si nu ma pricep la astea
Lg(x²)=-lg(2x-1)
Ajutor repede

Answer 1

a)Domeniul de   existenta a   ecuatiei
x+1>0 x>-1
(x+1)=3²
x+1=9
x=9-1=8
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b)Do,eniul  de  existenta   a  ecuatiei
x²≠0   x≠0
2x-1>0 x>1/2
Lgx²=lg(2x-1)^(-1)
lgx²=lg1/2x-1
Logaritmii  au   aceiasi  baza, poti delogaritma
x²=1/(2x-1)
x²(2x-1)=1
2x³-x²=1
2x³-x²-1=0
(x³-x²)+(x³-1)=0
x²(x-1)+(x-1)(x²+x+1)=0
(x-1)(x²+x²+x+1)=0
(x-1)*(2x²+x+1)=0
x-1=0   x1=1
2x²+x+1=0 Δ=1-8=-7<0  ecuatia  nu   admite  solutii   reale=>
x=1