Question

log in baza 2 din (x-1) = log in baza 4 din (3-x)

Answer 1

$log_{2} (x-1)=log_{4}(3-x)\\  [tex]log_{2}(x-1)=log_{2^{2} }(3-x)\\ log_{2}(x-1)=\frac{1}{2}log_{2}  (3-x)\\    x-1=\sqrt{3-x} \\ DVA:x>1\\ (x-1)^2=3-x\\ x^{2} -2x+1=3-x\\ x^{2} -x-2=0\\ x=-1 sau x=2\\ Raspuns:x=2$