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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)

User Nubkadiya
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1 Answer

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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-(r)/(100))^x

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


30000=52000(1-(12.5)/(100))^x


30000=52000((100)/(100)-(12.5)/(100))^x


30000=52000((100-12.5)/(100))^x


30000=52000((87.5)/(100))^x


30000=52000(0.875)^x

Dividing both sides by 52000, we get


(30000)/(52000)=(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=(\log _bx)/(\log _ba)

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


\log _(0.875)0.577=(\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=(-0.238)/(-0.058)=4.103

Taking the nearest whole number.


\log _(0.875)0.577=4

Hence the number of years =4.

User Leon Overweel
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