Boa noite! Vamos calcular o período 'n' com os dados: \\C=17400\\ J=13318,38\\ M=C+J=17400+13318,38=30718,38\\ i=3,4\%\text{ a.m.}\\ M=C(1+i)^n\\ 30718,38=17400(1+3,4\%)^n\\ 30718,38=17400(1+0,034)^n\\ 1,034^n=\frac{30718,38}{17400}\\ n\approx \frac{\log{1,765424}}{\log{1,034...