Respuesta :
We have been given that the value of Vishal's car is depreciating exponentially. Â The relationship between V, the value of his car, in dollars, and t, the elapsed time, in years, since he purchased the car is modeled by the equation [tex]V=22,500\cdot 10^{-\frac{t}{12}}[/tex]. We are asked to find the time, when Vishal's car will be worth $10,000.
To find the time, we will equate our given equation with 10,000 as:
[tex]10,000=22,500\cdot 10^{-\frac{t}{12}}[/tex] Â
[tex]\frac{10,000}{22,500}=\frac{22,500\cdot 10^{-\frac{t}{12}}}{22,500}[/tex]
[tex]\frac{4}{9}=10^{-\frac{t}{12}}[/tex]
Now we will take log on both sides.
[tex]log_{10}\left(\frac{4}{9}\right)=log_{10}\left(10^{-\frac{t}{12}}\right)[/tex]
[tex]log_{10}\left((\frac{2}{3})^2\right)=-\frac{t}{12}[/tex]
[tex]-\frac{t}{12}=log_{10}\left((\frac{2}{3})^2\right)[/tex]
[tex]-\frac{t}{12}(-12)=(-12)log_{10}\left((\frac{2}{3})^2\right)[/tex]
[tex]t=(-12)log_{10}\left((\frac{2}{3})^2\right)[/tex]
Using property [tex]log(a^b)=b\cdot log(a)[/tex], we will get:
[tex]t=(-12)\cdot 2log_{10}\left(\frac{2}{3}\right)[/tex]
[tex]t=-24log_{10}\left(\frac{2}{3}\right)[/tex]
Therefore, the value of Vishal's car will be $10,000 after approximately  [tex]-24log_{10}\left(\frac{2}{3}\right)[/tex] years.
