A piece of machinery valued at $20,000 depreciates at a rate of 9% yearly. How long will it take until it has a value of $10,000? (Express your answer to the nearest tenth.) its not 7.4
A = P * (1 - r)^t 10000 = 20000*(.91)^t 1/2 = (.91)^t t = log(1/2)/log(.91) t = 7.3 years
Are you sure, I got 7.34961458... Just plug in 20000*.91^x=10000 which is .91^x=1/2 which means you put in log(1/2)/log(.91)=x and you should get 7.34961458...
Depreciation is straight line so it is $20,000 * .09 = $1800 / yr (20,000 - 10,000) / 1800 = 5.6 years
20000 / (1.09)^x = 10000 x = 8.043 years Remember 100 * 1.09 * 0.91 = 99.19 ... NOT 100
