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Consider the function $$F(x) = 1-\frac{200^{ 2.5 }}{ x^{2.5}}.$$ We want to solve $F(x) = 0.3$.

  1. Since $ F(x) = 0.3$ then we can say, $2.5 \ln (\frac{200}{x}) = 0.7$.

  2. $\ln(\frac{200}{x}) = 0 .28$.

  3. $e^{\ln(\frac{200}{x})} = e^{.28}$.

  4. $\frac{200}{x} = e^{.28}$.

  5. Hence $x = \frac{200}{e^{.28}}$

  6. The answer appears to be about $x = 151.16$.

Just to sum it up, the actual answer is $230.66$, and this was achieved by taking the 2.5th root instead of doing the natural log. Why wasn't it the same answer?

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Because it should be $\ln 0.7$ on the RHS, not $0.7$.

$ F(x) = 0.3 \Rightarrow\\ \dfrac{200^{2.5}}{x^{2.5}} = 1 - 0.3 = 0.7 \Rightarrow\\ 2.5\ln \dfrac{200}{x} = \ln 0.7 \Rightarrow\\ \ln \dfrac{200}{x} = -0.14267\Rightarrow\\ \dfrac{200}{x} = e^{-0.14267} \Rightarrow\\ x = 200e^{0.14267} = \boxed{230.67} $

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  • $\begingroup$ Of course, here I deliberately calculated $\frac{\ln 0.7}{2.5}$ for otherwise, we'd end up getting $0.7^{1/2.5}$, which you want to avoid. $\endgroup$ – M. Vinay Jun 11 '14 at 6:48

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