Zadatak log-001

Oblast definisanosti: $x>0,\ x\ne 1$. Po formuli prelaska na drugu osnovu $\log_{x}2=\dfrac{1}{\log_{2}x}$. Smena $t=\log_{2}x$ ($t\ne 0$):

$t+\frac{1}{t}=\frac{5}{2}\;\Rightarrow\;2t^{2}-5t+2=0\;\Rightarrow\;(2t-1)(t-2)=0\;\Rightarrow\;t=2\ \text{ili}\ t=\frac12.$

Tada $\log_{2}x=2\Rightarrow x=4$ i $\log_{2}x=\dfrac12\Rightarrow x=\sqrt{2}$. Oba korena zadovoljavaju oblast definisanosti.