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| rev | Admin | (Created page with "'''Solution: A''' Equating the accumulated values after 4 years provides an equation in K <math display = "block"> \begin{align*} 10(1+\frac{K}{25})^{4} &= 10\exp \left(\int_{0}^{*4}\frac{1}{K+0.25t}d t\right) \\ 4\ln(1+0.04K) &= \int_{0}^{4}\frac{1}{K+0.25t}d t=4\ln(K)+0.25t)\big|_{0}^{4} \\ &=4\ln(\operatorname{K}+0.25t)\big|_{0}^{4} \\ &=4\ln(K+1)-4\ln(K) = 4\ln\frac{K+1}{K} \\ & 1+0.04K = \frac{K+1}{K}\\ & 0.04K^2 = 1 \\ &K = 5 \end{align*} </math> Therefore <ma...") | Nov 17'23 at 22:39 | +534 |