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Sagot :
Answer in terms of Ln: [tex]20*\text{Ln}\left(\frac{25}{13}\right)\\\\[/tex]
Answer in decimal form: 13.1
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Work Shown:
[tex]P(t) = 1,300,000e^{0.05t}\\\\2,500,000 = 1,300,000e^{0.05t}\\\\e^{0.05t} = \frac{2,500,000}{1,300,000}\\\\e^{0.05t} = \frac{25}{13}\\\\0.05t = \text{Ln}\left(\frac{25}{13}\right)\\\\t = 20*\text{Ln}\left(\frac{25}{13}\right)\\\\t \approx 13.0785293481332\\\\t \approx 13.1\\\\[/tex]
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