flatantsmu

2022-09-06

Solving $F\left(x\right)=0.3$ where $F\left(x\right)=1-\frac{{200}^{2.5}}{{x}^{2.5}}$
Consider the function
$F\left(x\right)=1-\frac{{200}^{2.5}}{{x}^{2.5}}.$
We want to solve $F\left(x\right)=0.3$
1.Since $F\left(x\right)=0.3$ then we can say, $2.5\mathrm{ln}\left(\frac{200}{x}\right)=0.7$
2.$\mathrm{ln}\left(\frac{200}{x}\right)=0.28$
3.${e}^{\mathrm{ln}\left(\frac{200}{x}\right)}={e}^{.28}$
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?

Mckenna Friedman

Because it should be $\mathrm{ln}0.7$ on the RHS, not $0.7$
$F\left(x\right)=0.3⇒\phantom{\rule{0ex}{0ex}}\frac{{200}^{2.5}}{{x}^{2.5}}=1-0.3=0.7⇒\phantom{\rule{0ex}{0ex}}2.5\mathrm{ln}\frac{200}{x}=\mathrm{ln}0.7⇒\phantom{\rule{0ex}{0ex}}\mathrm{ln}\frac{200}{x}=-0.14267⇒\phantom{\rule{0ex}{0ex}}\frac{200}{x}={e}^{-0.14267}⇒\phantom{\rule{0ex}{0ex}}x=200{e}^{0.14267}=\overline{)230.67}$

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