Question

REZOLVATI ECUATIA:
a) log in baza 2 din x + log in baza 3 din 6 =6
b) log in baza 2 (x^2 + x )=1
c) lg x supra
________
lg (5x -4 ) =1/2

Answer 1

$a) \log_2 x+\log _3 6=6\\\log_2 x=6-\log_3 6\\x=2^{6-\log_3 6}\\\\b)\log_2 (x^2+x)=1\\x^2+x=2\\x^2+x-2=0\\\Delta=1+8=9\Rightarrow \sqrt{\Delta}=3\\x_1= \dfrac{-1+3}{2}=1\\x_2=\dfrac{-1-3}{2}=-2\\\\c)\dfrac{\lg x}{\lg(5x-4)}=\dfrac{1}{2}\\\text{Conditii de existenta:} x> \dfra{4}{5}\\2\lg x=\lg(5x-4)\\\lg x^2=\lg(5x-4)\\\text{Din injectivitatea functiei logaritm obtinem:}\\x^2=5x-4\\x^2-5x+4=0\\\Delta=25-16=9\Rightarrow \sqrt{\Delta}=3\\x_1=\dfrac{5+3}{2}=4\\x_2=\dfrac{5-3}{2}=1$