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Sagot :
so, the car depreciates by half every 5 years, or namely it depreciates by 50% every 5 years, so
[tex]\textit{Periodic/Cyclical Exponential Decay} \\\\ A=P(1 - r)^{\frac{t}{c}}\qquad \begin{cases} A=\textit{current amount}\dotfill &\$7000\\ P=\textit{initial amount}\dotfill &\$19000\\ r=rate\to 50\%\to \frac{50}{100}\dotfill &0.5\\ t=\textit{elapsed time}\\ c=period\dotfill &5 \end{cases}[/tex]
[tex]7000=19000(1-0.5)^{\frac{t}{5}}\implies \cfrac{7000}{19000}=0.5^{\frac{t}{5}}\implies \cfrac{7}{19}=0.5^{\frac{t}{5}} \\\\\\ \log\left( \cfrac{7}{19} \right)=\log\left( 0.5^{\frac{t}{5}} \right)\implies \log\left( \cfrac{7}{19} \right)=t\log\left( 0.5^{\frac{1}{5}} \right) \\\\\\ \log\left( \cfrac{7}{19} \right)=t\log( \sqrt[5]{0.5})\implies \cfrac{\log\left( \frac{7}{19} \right)}{\log( \sqrt[5]{0.5})}=t\implies 7.2\approx t[/tex]
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