a car was bought for 45000 in 2005. if the car loses 4.5% of its value semiannually  how much will it be worth at the end of 2017? in what year will the car be worth 0

To solve this we are going to use the compounded interest formula: [tex]A=P(1+ \frac{r}{n} )^{nt}[/tex]
where 
[tex]A[/tex] is the final value of the car after [tex]t[/tex] years
[tex]P[/tex] is the price of the car
[tex]r[/tex] is the rate in decimal form
[tex]n[/tex] is the number of times the rate is compounded per year 
[tex]t[/tex] is the time in years

Part 1. Notice that since the car is losing 4.5% of its value semiannually, it is losing 9% annually; also the rate is going to be negative, so [tex]r= \frac{-9}{100} =-0.09[/tex]. Since the car its value semiannually, [tex]n=2[/tex]. Also, to find [tex]t[/tex], we are going to subtract 2005 from 2017: [tex]t=2017-2005=12[/tex]. Finally, we know for our problem that [tex]P=45000[/tex]. Now that we have all the vales we need, lest replace them in our formula:
[tex]A=P(1+ \frac{r}{n} )^{nt}[/tex]
[tex]A=45000(1+ \frac{-0.09}{2} )^{(2)(12)[/tex]
[tex]A=45000(0.955)^{24}[/tex]
[tex]A=14903.68[/tex]

We can conclude that at the end of 2017 the char will be worth $14,903.68

Part 2. Since we want to know the year in which the price of the car will be zero, [tex]A=0[/tex]. From our previous calculations we know that [tex]P=45000[/tex], [tex]r=-0.045[/tex],and [tex]n=2[/tex]. Lets replace those values in our formula one more time:
[tex]A=P(1+ \frac{r}{n} )^{nt}[/tex]
[tex]0=45000(1+ \frac{-0.045}{2} )^{2t}[/tex]
Since [tex]t[/tex] is the exponent, we are going to use logarithms to bring it down: 
[tex] \frac{45000}{0} =(1+ \frac{-0.045}{2} )^{2t}[/tex]
[tex](1+ \frac{-0.045}{2} )^{2t}=0[/tex]
[tex]0.9775^{2t}=0[/tex]
[tex]ln(0.9775^{2t})=ln(0)[/tex]

Since [tex]ln(0)[/tex] cannot be evaluated (you can't divide by zero in mathematics), we can conclude that the value of the car will never be zero.