B)
F(t) is continuous for a given value of t if, and only if, the limit for F(t) for t approaching this values by the left and by the right result in the same value. To check it, let us to calculate it, as follows:
[tex]\begin{gathered} lim_{t\rightarrow4^-}^\text{ \lparen}5\cdot2^{t+1}-3)=5\cdot2^{4+1}-3= \\ \\ =5\cdot2^5-3=5\cdot32-3=160-3=157 \end{gathered}[/tex][tex]\begin{gathered} lim_{t\rightarrow4^{^+}}(\frac{500t-1500}{t-1})=\frac{500\cdot4-1500}{4-1}= \\ \\ \frac{2000-1500}{3}=\frac{500}{3}\cong166.7 \end{gathered}[/tex]
Because the calculation of both side-approaching limits results in different values, we are able to conclude that: The function is not continuous.
C)
To the given equation, we just need to substitute M(t) = 100 and isolate t to find its value, as follows:
[tex]\begin{gathered} M(t)=100=\frac{210(2^t-1)}{2^t+5}\Rightarrow10(2^t+5)=21(2^t-1) \\ \\ \Rightarrow10\cdot2^t+50=21\cdot2^t-20\Rightarrow21\cdot2^t-10\cdot2^t=50+20 \\ \Rightarrow11\cdot2^t=70\Rightarrow2^t=\frac{70}{11}\Rightarrow ln(2^t)=ln(\frac{70}{11}) \\ \Rightarrow t\cdot ln(2)=ln(\frac{70}{11})\Rightarrow t=\frac{ln(\frac{70}{11})}{ln(2)} \\ \Rightarrow t\cong2.67 \end{gathered}[/tex]
From the solution developed above, we are able to conclude that the solution for the given question is:
t = 2.67
