Harison ,
a)
[tex3]\text{T(t)}=2^{t}+400.2^{-t}[/tex3]
[tex3]\text{T(0)}=2^{0}+400.2^{-0}[/tex3]
[tex3]\text{T(0)}=1+400[/tex3]
[tex3]\text{T(0)}=401 \text{ °C}[/tex3]
[tex3]\text{T(1)}=2^{1}+400.2^{-1}[/tex3]
[tex3]\text{T(1)}=2+200[/tex3]
[tex3]\text{T(1)}=202\text{ °C}[/tex3]
b)
[tex3]40=2^{t}+400.2^{-t}[/tex3]
[tex3]40=2^{t}+\frac{400}{2^{t}}[/tex3]
Trocando [tex3]2^t=a[/tex3]
[tex3]40=a+\frac{400}{a}[/tex3]
[tex3]a^{2}-40.a+400=0[/tex3]
[tex3]a=\frac{-(-40)\pm\sqrt{(-40)^{2}-4.1.400} }{2.1}[/tex3]
[tex3]a=\frac{40\pm\sqrt{1600-1600} }{2}[/tex3]
[tex3]a=20[/tex3]
Destrocando:
[tex3]2^t=a[/tex3]
[tex3]2^t=20[/tex3]
Aplicando logaritmo de ambos os lados:
[tex3]\log_22^t=\log_220[/tex3]
[tex3]t=\log_2(4.5)[/tex3]
[tex3]t=\log_24+\log_25[/tex3]
[tex3]t=2+2,3[/tex3]
[tex3]{\color{red}\boxed{t=4,3 \text{ h}}}[/tex3]