A car with a cost of $25,000 is decreasing in value at a rate of 10% each year. The function #g(t) = 25,000(.9)^t# gives the value of the car after #t# years. When will the value of the car be about $12,000?

1 Answer
Jim G.
Aug 29, 2017

#~~7" years"#

Explanation:

#"setting up the following equation"#

#25000(0.9)^t=12000#

#rArr(0.9)^t=12000/25000=12/25#

#"using the "color(blue)"law of logarithms"#

#•color(white)(x)logx^nhArrnlogx#

#"taking the ln (natural log ) of both sides"#

#ln(0.9)^t=ln(12/25)#

#rArrtln(0.9)=ln(12/25)#

#rArrt=(ln(12/25))/(ln(0.9))=6.966~~ 7" years"#

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