Como resolver?

\(\[\begin{align} & C\text{ }=\text{ }50.000 \\ & J\text{ }=\text{ }1.000 \\ & i\text{ }=\text{ } \\ & n\text{ }=\text{ } \\ & Equacionando: \\ & J=C*[{{(1+i)}^{n}}-1] \\ & Substituindo: \\ & 1.000=50.000*[{{(1+i)}^{n}}-1] \\ & \frac{1.000}{50.000}+1={{(1+i)}^{n}} \\ & 1,02={{(1+i)}^{n}}\to (1) \\ \end{align}\] \)

\(\[\begin{align} & Segunda\text{ equacao:} \\ & n\text{ }+\text{ }2 \\ & Substituindo: \\ & J\text{ }+\text{ }2.060,40 \\ & 1.000\text{ }+\text{ }2.060,40\text{ }=\text{ }3.060,40 \\ & 3.060,40=50.000*[{{(1+i)}^{(n+2)}}-1] \\ & \frac{3.060,40}{50.000}+1={{(1+i)}^{(n+2)}} \\ & 1,061208={{(1+i)}^{(n+2)}}\to (2) \\ \end{align}\] \)

\(\[\begin{align} & Temos: \\ & \\ & Dividindo\frac{\left( 2 \right)}{\left( 1 \right)}: \\ & \frac{1,061208}{1,02}=\frac{{{(1+i)}^{(n+2)}}}{{{(1+i)}^{n}}} \\ & 1,0404={{(1+i)}^{2}} \\ & {{1,0404}^{(1/2)}}=1+i \\ & 1,02\text{ }=\text{ }1\text{ }+\text{ }i \\ & i\text{ }=\text{ }1,02\text{ }-1 \\ & i\text{ }=\text{ }0,02\to \frac{2}{100}\to 2%a.m \\ & Para: \\ & i\text{ }=\text{ }0,02\text{ }em\text{ }\left( 1 \right) \\ & 1,02={{(1+i)}^{n}} \\ & 1,02={{(1+0,02)}^{n}} \\ & 1,02={{1,02}^{n}} \\ & n\text{ }=\text{ }\frac{log\text{ }1,02\text{ }}{log\text{ }1,02} \\ & n\text{ }=\text{ }1\text{ m }\!\!\hat{\mathrm{e}}\!\!\text{ s} \\ \end{align}\] \)