Admita que o n\u00famero de habitantes de um certo pa\u00eds \u00e9 dado por:<\/p>\n
$$N(t)=\\frac{100}{1+9\\times {{e}^{-0,18\\,t}}}$$<\/p>\n
com $N$ expresso em milh\u00f5es e sendo $t$ o n\u00famero de anos contados desde o in\u00edcio do ano 2000.<\/p><\/blockquote>\n
\u00ad<\/p>\n
\n- \u00a0Como $$\\begin{array}{*{35}{l}}
\nN(0) & = & \\frac{100}{1+9\\times {{e}^{0}}}\u00a0 \\\\
\n{} & = & \\frac{100}{10}\u00a0 \\\\
\n{} & = & 10\u00a0 \\\\
\n\\end{array}$$
\nem 2000, o n\u00famero de habitantes era 10 milh\u00f5es.
\n\u00ad<\/li>\n - Ora, $$\\begin{array}{*{35}{l}}
\nN(t)=2\\times N(0) & \\Leftrightarrow\u00a0 & \\frac{100}{1+9\\times {{e}^{-0,18\\,t}}}=2\\times 10\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & 1+9\\times {{e}^{-0,18\\,t}}=5\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & {{e}^{-0,18\\,t}}=\\frac{4}{9}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & -0,18\\,t=\\ln \\frac{4}{9}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & t=-\\frac{50}{9}\\ln \\frac{4}{9}\u00a0 \\\\
\n\\end{array}$$
\nComo $-\\frac{50}{9}\\ln \\frac{4}{9}\\approx 4,505$, a popula\u00e7\u00e3o duplicou no m\u00eas de julho de 2004.
\n\u00ad<\/li>\n - Ora, $$\\begin{array}{*{35}{l}}
\nN(t)=45 & \\Leftrightarrow\u00a0 & \\frac{100}{1+9\\times {{e}^{-0,18\\,t}}}=45\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & 1+9\\times {{e}^{-0,18\\,t}}=\\frac{20}{9}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & {{e}^{-0,18\\,t}}=\\frac{11}{81}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & -0,18\\,t=\\ln \\frac{11}{81}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & t=-\\frac{50}{9}\\ln \\frac{11}{81}\u00a0 \\\\
\n\\end{array}$$
\nComo $-\\frac{50}{9}\\ln \\frac{11}{81}\\approx 11,092$, conclui-se que ser\u00e3o atingidos 45 milh\u00f5es de habitantes no ano de 2011.
\n\u00ad<\/li>\n - Se $t\\to +\\infty $, ent\u00e3o ${{e}^{-0,18\\,t}}\\to 0$.Logo, se $t\\to +\\infty $, ent\u00e3o $N(t)=\\frac{100}{1+9\\times {{e}^{-0,18\\,t}}}\\to \\frac{100}{1}=100$.\n
Portanto, continuando v\u00e1lido o modelo e a longo prazo, o n\u00famero de habitantes ser\u00e1 pr\u00f3ximo de 100 milh\u00f5es.
\n\u00ad<\/li>\n<\/ol>\n