Respuesta :
a)
[tex]\bf 1990-1910=80\leftarrow t \\\\\\ P(t)=8000(2)^{-\frac{t}{29}}\implies P(80)=8000(2)^{-\frac{80}{29}} \\\\\\ P(80)=8000\cdot \cfrac{1}{2^{\frac{80}{29}}}\implies P(80)=\cfrac{8000}{\sqrt[29]{2^{80}}}[/tex]
and surely you know how much that is.
b)
[tex]\bf P(t)=125\\\\\\ 125=8000(2)^{-\frac{t}{29}}\implies \cfrac{125}{8000}=2^{-\frac{t}{29}}\implies \cfrac{1}{64}=2^{-\frac{t}{29}} \\\\\\ \textit{now we take log to both sides} \\\\\\ log\left( \frac{1}{64} \right)=log\left( 2^{-\frac{t}{29}} \right)\implies log\left( \frac{1}{64} \right)=-\frac{t}{29}log\left( 2 \right) \\\\\\ log\left( \cfrac{1}{64} \right)=-\cfrac{tlog(2)}{29}\implies \cfrac{-29log\left( \frac{1}{64} \right)}{log(2)}=t\implies 174=t[/tex]
since in 1910 t = 0, 174 years later from 1910, is 2084, so in 2084 they'll be 125 exactly, so the next year, 2085, will then be the first year they'd fall under that.
[tex]\bf 1990-1910=80\leftarrow t \\\\\\ P(t)=8000(2)^{-\frac{t}{29}}\implies P(80)=8000(2)^{-\frac{80}{29}} \\\\\\ P(80)=8000\cdot \cfrac{1}{2^{\frac{80}{29}}}\implies P(80)=\cfrac{8000}{\sqrt[29]{2^{80}}}[/tex]
and surely you know how much that is.
b)
[tex]\bf P(t)=125\\\\\\ 125=8000(2)^{-\frac{t}{29}}\implies \cfrac{125}{8000}=2^{-\frac{t}{29}}\implies \cfrac{1}{64}=2^{-\frac{t}{29}} \\\\\\ \textit{now we take log to both sides} \\\\\\ log\left( \frac{1}{64} \right)=log\left( 2^{-\frac{t}{29}} \right)\implies log\left( \frac{1}{64} \right)=-\frac{t}{29}log\left( 2 \right) \\\\\\ log\left( \cfrac{1}{64} \right)=-\cfrac{tlog(2)}{29}\implies \cfrac{-29log\left( \frac{1}{64} \right)}{log(2)}=t\implies 174=t[/tex]
since in 1910 t = 0, 174 years later from 1910, is 2084, so in 2084 they'll be 125 exactly, so the next year, 2085, will then be the first year they'd fall under that.
