Question

log baza 2 dinx +log baza x di2=2

Answer 1

$log_{2}x+ log_{x}2 =2;x\ \textgreater \ 0;x \neq 1 \\ log_{2}x=t \\ t+ \frac{1}{t}=2 \\ t^{2}-2t+1=0 \\ (t-1)^2=0 \\ t=1 \\ log_{2}x=1 \\ x=2$

Answer 2

Conditii de existenta: x>0 si x diferit de 1.

log_2 (x) + log_x (2) = 2 <=> log_2 (x) +1/(log_2 (x)) = 2.
Notam t=log_2 (x) si ecuatia devine t+1/t = 2 <=> t^2+1 = 2t <=> t^2-2t+1 = 0 <=> (t-1)^2 = 0 <=> t = 1 <=> log_2 (x) = 1 <=> x=2^1=2.