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Sagot :
Using the inverse function, it is found that the equation is:
[tex]P(t) = \frac{e^{0.1t}}{50} + 20[/tex]
It's initial value is 20.
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We want to find the inverse of:
[tex]y = P^{-1}(t) = 10\ln{50t - 1000}[/tex]
To do this, we exchange y and t, and isolate y.
Then:
[tex]10\ln{50y - 1000} = t[/tex]
[tex]\ln{50y - 1000} = \frac{t}{10}[/tex]
[tex]e^{\ln{50y - 1000}} = e^{0.1t}[/tex]
[tex]50y - 1000 = e^{0.1t}[/tex]
[tex]50y = e^{0.1t} + 1000[/tex]
[tex]y = \frac{e^{0.1t} + 1000}{50}[/tex]
[tex]y = \frac{e^{0.1t}}{50} + \frac{1000}{50}[/tex]
[tex]P(t) = \frac{e^{0.1t}}{50} + 20[/tex]
It's initial value is:
[tex]P(0) = \frac{e^{0.1(0)}}{50} + 20 = 0.02 + 20 = 20.02[/tex]
Rounding, 20.
A similar problem is given at https://brainly.com/question/23950969
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