Question

The relationship between V, the value of his car, in dollars, and t, the elapsed time, in years, since he

Answer 1

Answer:

The number of years after purchase at which Vishal's car will be worth $10,000 is $Log_{10}\left (\frac{9}{4} \right )^{12}$  years

Step-by-step explanation:

The relationship is given as follows

Value, V of the car = 22500×$10^{-t/12}$

Therefore, when the car is $10,000 we will have

$10,000 = 22,500×$10^{-t/12}$

Which will give;

$\frac{10000}{22500} =\frac{4}{9} = 10^{-\frac{t}{12}}$

Hence;

$Log\frac{4}{9} = Log(10^{-\frac{t}{12}} )$

Therefore;

$-\frac{t}{12}\times Log_{10} 10 = Log\frac{4}{9}$

$\because Log_{10} 10 = 1, \ we \ have; \ -\frac{t}{12} = Log_{10}\frac{4}{9}$

Which gives;

$t = -12 \times Log_{10}\frac{4}{9} \ or \ t = Log_{10}\left (\frac{4}{9} \right )^{-12}$

$\therefore t = Log_{10}\left (\frac{9}{4} \right )^{12}$ years

Evaluated, the above equation becomes t = 4.226 years

Therefore, the number of years after purchase at which Vishal's car will be worth $10,000 = $Log_{10}\left (\frac{9}{4} \right )^{12}$  years.

Answer 2

Answer:

t = -24*log(2/3)

Step-by-step explanation:

The expression is:

V = 22,500*10^(-t/12)

Replacing with V = 10,000 and isolating t, we get:

10,000 = 22,500*10^(-t/12)

10,000/22,500 = 10^(-t/12)

4/9 = 10^(-t/12)

(2/3)² = 10^(-t/12)

2*log(2/3) = -t/12

-12*2*log(2/3) = t

t = -24*log(2/3)