The value of Vishal's car is depreciating exponentially. $\newline$ 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 following equation. $\newline$ $V=22,500 \cdot 10^{-\frac{t}{12}}$ $\newline$ How many years after purchase will Vishal's car be worth $\$ 10,000$ ? $\newline$ Give an exact answer expressed as a base $-10$ logarithm. $\newline$ years

Full solution

Q. The value of Vishal's car is depreciating exponentially.

\newline

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 following equation.

\newline

V=22,500 \cdot 10^{-\frac{t}{12}}

\newline

How many years after purchase will Vishal's car be worth

\$ 10,000

\newline

Give an exact answer expressed as a base

-10

logarithm.

\newline

years

Set up equation: Set up the equation with the given value of the car. $\newline$ We are given the equation $V = 22,500 \times 10^{-\frac{t}{12}}$ and we want to find the time $t$ when $V = \$10,000$ . $\newline$ So, we set up the equation $10,000 = 22,500 \times 10^{-\frac{t}{12}}$ .
Divide sides: Divide both sides of the equation by $22,500$ to isolate the exponential term. $\newline$ $10,000 / 22,500 = 10^{-(t)/(12)}$ $\newline$ This simplifies to $4/9 = 10^{-(t)/(12)}$ .
Convert to logarithmic form: Convert the equation to logarithmic form to solve for $t$ . We can take the base $-10$ logarithm of both sides to get $\log(\frac{4}{9}) = \log(10^{-\frac{t}{12}})$ . Using the property of logarithms that $\log(b^x) = x \cdot \log(b)$ , we get $\log(\frac{4}{9}) = -\frac{t}{12} \cdot \log(10)$ . Since $\log(10)$ is $1$ , this simplifies to $\log(\frac{4}{9}) = -\frac{t}{12}$ .
Solve for t: Solve for t. $\newline$ To isolate $t$ , we multiply both sides by $-12$ : $-12 \times \log(\frac{4}{9}) = t$ .
Calculate exact value: Calculate the exact value of $t$ . $t = -12 \times \log(\frac{4}{9})$ .This is the exact answer expressed as a base- $10$ logarithm.