The more accurate value is 6.57881347896059, which you can round however you need. I picked two decimal places.
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Explanation:
Let's pick a starting value of the car. It doesn't matter what the starting value, but it might help make the problem easier. Let's say A = 1000. Half of that is 1000/2 = 500.
So we want to find out how long it takes for the car's value to go from $1000 to $500 if it depreciates 10% per year.
The value of r is r = 0.10 as its the decimal form of 10%
t is the unknown number of years we want to solve for
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y = A(1 - r)^t
500 = 1000(1 - 0.1)^t
500 = 1000(0.9)^t
1000(0.9)^t = 500
0.9^t = 500/1000
0.9^t = 0.5
log( 0.9^t ) = log( 0.5 )
t*log( 0.9 ) = log( 0.5 )
t = log( 0.5 )/log( 0.9 )
t = 6.57881347896059
Note the use of logs to help us isolate the exponent.
