Question

Kevin buys a truck for $52,000 and it decreases in value 12.5% every year, How many years will it take for the truck to be worth $30,000 (Use log method, nearest whole number)

Answer 1

Given :

Principal P= $52000

Interest rate r=12.5 %

Let x be the number of yers

The worth p(x) is $30000 after x years.

The math model for worth is

$P(x)=P(1-\frac{r}{100})^x$

Substitute P(x)=30000, P=52000, r=12.5, we get

$30000=52000(1-\frac{12.5}{100})^x$

$30000=52000(\frac{100}{100}-\frac{12.5}{100})^x$

$30000=52000(\frac{100-12.5}{100})^x$

$30000=52000(\frac{87.5}{100})^x$

$30000=52000(0.875)^x$

Dividing both sides by 52000, we get

$\frac{30000}{52000}=\frac{52000}{52000}(0.875)^x$

$0.577=0.875^x$

We know that

$b^x=y\text{ then }log_by=x$

Here b=0.875 and y=0.577.

$\log _{0.875}0.577=y$

Using the change base formula.

$\log _ax=\frac{\log _bx}{\log _ba}$

Take b=10, and substitute a=0.875 and x=0.577, we get

$\log _{0.875}0.577=\frac{\log _{10}0.577}{\log _{10}0.875}$

$\text{ Use }\log _{10}0.577=-0.238\text{ and }\log 0.875=-0.058.$

$\log _{0.875}0.577=\frac{-0.238}{-0.058}=4.103$

Taking the nearest whole number.

$\log _{0.875}0.577=4$

Hence the number of years =4.