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Question Number 206302 by ayietaamos last updated on 11/Apr/24

Answered by A5T last updated on 11/Apr/24

$$k=\frac{ln\left(\frac{T}{3y}\right)-1}{2}$$

Answered by MATHEMATICSAM last updated on 11/Apr/24

$$T=3{ye}^{2k+1}$$$$\Rightarrow e^{2k+1}=\frac{T}{3y}$$$$\Rightarrow 2k+1=\ln \left(\frac{T}{3y}\right)$$$$\Rightarrow 2k=\ln \left(\frac{T}{3y}\right)-1$$$$\Rightarrow 2k=\ln \left(\frac{T}{3y}\right)-\ln e$$$$\Rightarrow 2k=\ln \left(\frac{T}{3ye}\right)$$$$\Rightarrow k=\frac{1}{2}\ln \left(\frac{T}{3ye}\right)$$

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