Car value

The car loses value 15% every year.

Determine a time (in years) when the price will be halved.

Correct result:

n =  4.265

Solution:

(1−15100)n=0.5 nln⁡(1−15100)=ln⁡0.5 n=ln⁡0.5ln⁡(1−15100) n=4.265 (1-\dfrac{15}{100})^n = 0.5 \ \\ n \ln (1-\dfrac{15}{100}) = \ln 0.5 \ \\ n = \dfrac{ \ln 0.5 }{ \ln (1-\dfrac{15}{100})} \ \\ n = 4.265 \ \\

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