Question

aveti mai jos exercitiul. chiar nu stiu ce sa ii fac :)). nu cred ca pot sa ii dau valori lui x, sau sa il urc sus, deci astept solutii :D.

Answer 1

ca lg sa aiba sens trebuie ca x>0
notam lgx=t si obtinem

t²-5t+6=0
t²-2t-3t+6=0
t(t-2)-3(t-2)=0
(t-2)(t-3)=0

t₁=2>0 => lgx=2 => x=10² => x=100
t₂=3>0 => lgx=3 => x=10³ => x=1000

O seara buna!

Answer 2

$\displaystyle lg^2x-5lgx+6=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x\ \textgreater \ 0~\\ \\ lgx=t;~lg^2x=(lgx)^2=t^2\\ \\ t^2-5t+6=0\\ \\ a=1,~b=-5,~c=6 \\ \\ \Delta=(-5)^2-4 \cdot 1 \cdot 6=25-24=1\ \textgreater \ 0 \\ \\ t_1 = \frac{5-1}{2} = \frac{4}{2} =2 \\ \\ t_2= \frac{5+1}{2} = \frac{6}{2} =3 \\ \\ lgx=2 \Rightarrow lgx=lg100 \Rightarrow x=100 \\ \\ lgx=3 \Rightarrow lgx=lg1000 \Rightarrow x=1000$