A hay baker used by a local farmer originally cost $22,000 depreciates at a constant rate of 8.5% per year. In how many years will the piece of equipment be worth only $8,500?

In 11 years

Step-by-step explanation:

To calculate the number of years it will take to depreciate to that value, we use the equation below; The amount after t years, with initial amount I at percentage of depreciation d can be represented as follows;

A = I( 1 - d)^t

In the question Using our definition, A = $8,500, I = $22,000, d = 8.5% = 8.5/100 = 0.085 and t = ?

Let’s plug these values;

8,500 = 22,000(1 - 0.085)^t

8,500 = 22,000(0.915)^t

divide both side by 22,000

8,500/22,000 = (0.915)^t

0.3864 = (0.915)^t

Taking the logarithm of both sides

log 0.3864 = log(0.915)^t

log 0.3864 = tlog 0.915

t = log 0.3864/log 0.915

t = 10.7 approximately 11 years

could

step-by-step explanation:

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