Zauważmy, że ${\log }_{a}b=\frac{1}{{\log }_{b}a}$, stąd:

${\log }_{a}b·{\log }_{b}a=\frac{1}{{\log }_{b}a}·{\log }_{b}a=1$.

$\sqrt{{\log }_{a}b}=\frac{{\log }_{a}b}{\sqrt{{\log }_{a}b}}$

Stąd:

$a^{\sqrt{{\log }_{a}b}}=b^{\frac{1}{\sqrt{{\log }_{a}b}}}=b^{\sqrt{{\log }_{b}a}}$.

Zatem:

$a^{\sqrt{{\log }_{a}b}}+\sqrt{{\log }_{a}b}\cdot \sqrt{{\log }_{b}a}-b^{\sqrt{{\log }_{b}a}}=0+1=1$.

Zapisujemy ${\log }_{2}x$ za pomocą logarytmu o podstawie $5$:

${\log }_{5}x·\frac{{\log }_{5}x}{{\log }_{5}2}={\log }_{5}2$

${\left({\log }_{5}x\right)}^2={\left({\log }_{5}2\right)}^2$.

Stąd:

${\log }_{5}x={\log }_{5}2$ lub ${\log }_{5}x=-{\log }_{5}2$

$x=2$ lub $x=0,5$.