honhon,
[tex3]\mathsf{
log(1,08) = log(\frac{108}{100}) = log(\frac{27}{25})=log27 - log 25 =\\
log(3)^3-log5^2 = 3log3 - 2log5 = 3log3 - 2(log(\frac{10}{2}))=\\
3log3 - 2log10 +2log2 = \boxed{3log3-2+2log2}\checkmark\\
\therefore log3 +log2=t(3log3-2+2log2)\implies (0,48+0,30) =t (1,44-2+0,6)\\
t = \frac{0,78}{0,04}=\boxed{19,5}\color{green}\checkmark
}[/tex3]...