Question

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

Answer 1

To solve this we are going to use the compounded interest formula: $A=P(1+ \frac{r}{n} )^{nt}$
where 
$A$ is the final value of the car after $t$ years
$P$ is the price of the car
$r$ is the rate in decimal form
$n$ is the number of times the rate is compounded per year 
$t$ 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 $r= \frac{-9}{100} =-0.09$. Since the car its value semiannually, $n=2$. Also, to find $t$, we are going to subtract 2005 from 2017: $t=2017-2005=12$. Finally, we know for our problem that $P=45000$. Now that we have all the vales we need, lest replace them in our formula:
$A=P(1+ \frac{r}{n} )^{nt}$
$A=45000(1+ \frac{-0.09}{2} )^{(2)(12)$
$A=45000(0.955)^{24}$
$A=14903.68$

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, $A=0$. From our previous calculations we know that $P=45000$, $r=-0.045$,and $n=2$. Lets replace those values in our formula one more time:
$A=P(1+ \frac{r}{n} )^{nt}$
$0=45000(1+ \frac{-0.045}{2} )^{2t}$
Since $t$ is the exponent, we are going to use logarithms to bring it down: 
$\frac{45000}{0} =(1+ \frac{-0.045}{2} )^{2t}$
$(1+ \frac{-0.045}{2} )^{2t}=0$
$0.9775^{2t}=0$
$ln(0.9775^{2t})=ln(0)$

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