Wiemy, że dla $a>1$, $b>1$ i $N>1$:

${\log }_{a^{2}b}N=\frac{3}{20}\cdot \left({\log }_{a}N+{\log }_{b}N\right)$

Chcemy wyznaczyć ${\log }_{a}b$.

Korzystamy ze wzoru na zmianę podstawy logarytmu:

${\log }_{a^{2}b}N=\frac{{\log }_{a}N}{{\log }_a{a}^{2}b}=\frac{{\log }_{a}N}{{\log }_a\left(a^2\cdot b\right)}=\frac{{\log }_{a}N}{{\log }_a{a}^2+{\log }_{a}b}=\frac{{\log }_{a}N}{2+{\log }_{a}b}$

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

Otrzymujemy, że:

$\frac{{\log }_{a}N}{2+{\log }_{a}b}=\frac{3}{20}\cdot \left({\log }_{a}N+\frac{{\log }_{a}N}{{\log }_{a}b}\right)$

$\frac{{\log }_{a}N}{2+{\log }_{a}b}=\frac{3}{20}\cdot {\log }_{a}N\cdot \left(1+\frac{1}{{\log }_{a}b}\right)|:{\log }_{a}N,{\log }_{a}N\ne 0(\text{bo }N\ne 1)$

$\frac{1}{2+{\log }_{a}b}=\frac{3}{20}\cdot \left(1+\frac{1}{{\log }_{a}b}\right)|\cdot 20$

$\frac{20}{2+{\log }_{a}b}=3\cdot \left(1+\frac{1}{{\log }_{a}b}\right)$

$\frac{20}{2+{\log }_{a}b}=3+\frac{3}{{\log }_{a}b}$

Podstawiamy:

$t={\log }_{a}b$

Wtedy:

$\frac{20}{2+t}=3+\frac{3}{t}$

$\frac{20}{2+t}-3-\frac{3}{t}=0$

$\frac{20t}{t\left(2+t\right)}-\frac{3t\left(2+t\right)}{t\left(2+t\right)}-\frac{3\left(2+t\right)}{t\left(2+t\right)}=0$

$\frac{20t-3t\left(2+t\right)-3\left(2+t\right)}{t\left(2+t\right)}=0|\cdot t\left(2+t\right),t\ne 0\text{i}t\ne -2$

$20t-3t\left(2+t\right)-3\left(2+t\right)=0$

$20t-6t-3t^2-6-3t=0$

$-3t^2+11t-6=0$

$\Delta =11^2-4\cdot \left(-3\right)\cdot \left(-6\right)=121-72=49$

$\sqrt{\Delta }=\sqrt{49}=7$

$t_1=\frac{-11-7}{2\cdot \left(-3\right)}=\frac{-11-7}{-6}=\frac{-18}{-6}=3$

$t_2=\frac{-11+7}{2\cdot \left(-3\right)}=\frac{-4}{-6}=\frac{2}{3}$

Ostatecznie:

$\boxed{{\log }_{a}b=3\text{lub}{\log }_{a}b=\frac{2}{3}}$