Question

$N(t)= 9^{t+1} -4 . 3^{t+1} +5$
em que n(t)=50 000

Answer 1

$50000~=~9^{t+1}-4\,.\,3^{t+1}+5\\\\\\50000~=~\left(3^2\right)^{t+1}-4\,.\,3^{t+1}+5\\\\\\Utilizando~a~propriedade~da~potencia~de~potencia\\\\\\50000~=~\left(3^2\right)^{t+1}-4\,.\,3^{t+1}+5\\\\\\Utilizando~a~mesma~propriedade\\\\\\50000~=~\left(3^{t+1}\right)^{2}-4\,.\,\left(3^{t+1}\right)+5\\\\\\\left(3^{t+1}\right)^{2}-4\,.\,\left(3^{t+1}\right)+5-50000~=~0\\\\\\\boxed{\left(3^{t+1}\right)^{2}-4\,.\,\left(3^{t+1}\right)-49995~=~0}$

Observando cuidadosamente a equação, podemos ver um "padrão" de equação do 2° grau (a.x² + b.x + c = 0)

Vamos, então, fazer uma substituição para facilitar a visualização:

$3^{t+1}~=~x$

Substituindo na equação:

$x^{2}-4x-49995~=~0\\\\\\Aplicando~Bhaskara\\\\\\\Delta~=~(-4)^2-4.1.(-49995)~=~16+199980~=~\boxed{199996}\\\\\\x'~=~\frac{4+\sqrt{199996}}{2\,.\,1}~=~\frac{4+\sqrt{4~.~49999}}{2}~=~\frac{4+2\sqrt{49999}}{2}~=~\boxed{2+\sqrt{49999}}\\\\\\x''~=~\frac{4-\sqrt{199996}}{2\,.\,1}~=~\frac{4-\sqrt{4~.~49999}}{2}~=~\frac{4-2\sqrt{49999}}{2}~=~\boxed{2-\sqrt{49999}}$

Podemos agora voltar a substituição feita (para x' e x'') e determinar o valor de "t".

$3^{t+1}~=~x':\\\\\\3^{t+1}~=~2+\sqrt{49999}\\\\\\Aplicando~logaritmo~de~base~3~nos~dois~lados~da~equacao:\\\\\\log_{_3}3^{t+1}~=~log_{_3}\left(2+\sqrt{49999}\right)\\\\\\t+1~=~log_{_3}\left(2+\sqrt{49999}\right)\\\\\\\boxed{t~=~log_{_3}\left(2+\sqrt{49999}\right)~-~1}$

$Podemos~ainda~passar~o~\,''-1''~para~dentro~do~log~utilizando~a\\propriedade~do~logaritmo~do~quociente\\\\\\t~=~log_{_3}\left(2+\sqrt{49999}\right)~-~log_{_3}3\\\\\\\boxed{t~=~log_{_3}\left(\dfrac{2+\sqrt{49999}}{3}\right)}$

$3^{t+1}~=~x'':\\\\\\3^{t+1}~=~2-\sqrt{49999}\\\\\\Perceba~que~~2-\sqrt{49999}~\acute{e}~um~numero~negativo,~logo~\underline{esta~equacao}\\\underline{nao~tem~solucao~nos~Reais}$

Com isso, chegamos a resposta que   $t~=~log_{_3}\left(\dfrac{2+\sqrt{49999}}{3}\right)$