Try \( t = 12 \):

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• + 0.05(10) = 1.5, \quad (1.04)^{10} \approx 1.4802
• Try \( t = 11 \):
• + 0.55 = 1.55, \quad (1.04)^{11} \approx 1.5396
• + 0.6 = 1.6, \quad (1.04)^{12} \approx 1.6010
• At \( t = 12 \), both values are approximately equal.
• Therefore, it takes 12 years.

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