Questão 2 - 1ª Fase - ITA 2026

Simplificando as expressões:

${\log }_{45}13,72=0,6880\to {\log }_{45}\frac{7^3·2^2}{5^2·2^2}=0,6880\to {\log }_{45}\frac{7^3}{5^2}=0,6880$

${\log }_{45}{7}^3-{\log }_{45}{5}^2=0,6880\to$

$3{\log }_{45}7-2{\log }_{45}5=0,6880 \mathbf{\left(}\mathbf{I}\mathbf{\right)}$

Também:

${\log }_{45}6125=2,2908\to {\log }_{45}{7}^2·5^3=2,2908\to$

${\log }_{45}{7}^2+{\log }_{45}{5}^3=2,2908\to$

$2{\log }_{45}\left(7\right)+3{\log }_{45}\left(5\right)=2,2908 \left(II\right)$

De $\left(I\right)$ e $\left(II\right)$:

$\left\{\begin{array}{l}3{\log }_{45}7-2{\log }_{45}5=0,6880\\ 2{\log }_{45}7+3{\log }_{45}5=2,2908\end{array}\right.$

Resolvendo o sistema, encontra-se:

${\log }_{45}7=\boxed{0,5112}$