# Solving $F(x) = 0.3$ where $F(x) = 1-\frac{200^{ 2.5 }}{ x^{2.5}}$

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?

Because it should be $\ln 0.7$ on the RHS, not $0.7$.
$F(x) = 0.3 \Rightarrow\\ \dfrac{200^{2.5}}{x^{2.5}} = 1 - 0.3 = 0.7 \Rightarrow\\ 2.5\ln \dfrac{200}{x} = \ln 0.7 \Rightarrow\\ \ln \dfrac{200}{x} = -0.14267\Rightarrow\\ \dfrac{200}{x} = e^{-0.14267} \Rightarrow\\ x = 200e^{0.14267} = \boxed{230.67}$
• 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. – M. Vinay Jun 11 '14 at 6:48