a)
Notam cu t = ln ( x - 1 ) => t^2 - t - 2 = 0
Δ= 1 + 8 = 9
t1 = ( 1 + 3) / 2 = 2 => ln ( x - 1 )= 2 -> x - 1 = e^2 -> x = e ^ 2 + 1
t2 = ( 1 - 3) / 2= -1 => ln ( x - 1 ) =-1 -> x - 1 = e^(-1) -> x=e^(-1)+1
b)
[tex]log_3(x+1)+log_3(x+3)=1[/tex] <=>
Conditii de existenta a logaritmului:
[tex]x+1>0 -> x>-1 \\x+3>0 ->x>-3\\\\log_3((x+1)*(x+3))=1 => (x+1)*(x+3)=3 => x^{2}+4x+3=3 =>\\ x^{2}+4x=0 <=> x(x+4)=0 => x=0 /x=-4\\[/tex]
Dar x >-1 -> x nu poate fi -4 -> x=0
c)Am nota [tex]2^{x}=t[/tex] si asa s-ar rezolva daca ecuatia respectiva ar fi egala cu o valoare
