flatantsmu

2022-09-06

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, $2.5\mathrm{ln}(\frac{200}{x})=0.7$

2.$\mathrm{ln}(\frac{200}{x})=0.28$

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

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?

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, $2.5\mathrm{ln}(\frac{200}{x})=0.7$

2.$\mathrm{ln}(\frac{200}{x})=0.28$

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

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?

Mckenna Friedman

Beginner2022-09-07Added 10 answers

Because it should be $\mathrm{ln}0.7$ on the RHS, not $0.7$

$F(x)=0.3\Rightarrow \phantom{\rule{0ex}{0ex}}{\displaystyle \frac{{200}^{2.5}}{{x}^{2.5}}}=1-0.3=0.7\Rightarrow \phantom{\rule{0ex}{0ex}}2.5\mathrm{ln}{\displaystyle \frac{200}{x}}=\mathrm{ln}0.7\Rightarrow \phantom{\rule{0ex}{0ex}}\mathrm{ln}{\displaystyle \frac{200}{x}}=-0.14267\Rightarrow \phantom{\rule{0ex}{0ex}}{\displaystyle \frac{200}{x}}={e}^{-0.14267}\Rightarrow \phantom{\rule{0ex}{0ex}}x=200{e}^{0.14267}=\overline{){\displaystyle 230.67}}$

$F(x)=0.3\Rightarrow \phantom{\rule{0ex}{0ex}}{\displaystyle \frac{{200}^{2.5}}{{x}^{2.5}}}=1-0.3=0.7\Rightarrow \phantom{\rule{0ex}{0ex}}2.5\mathrm{ln}{\displaystyle \frac{200}{x}}=\mathrm{ln}0.7\Rightarrow \phantom{\rule{0ex}{0ex}}\mathrm{ln}{\displaystyle \frac{200}{x}}=-0.14267\Rightarrow \phantom{\rule{0ex}{0ex}}{\displaystyle \frac{200}{x}}={e}^{-0.14267}\Rightarrow \phantom{\rule{0ex}{0ex}}x=200{e}^{0.14267}=\overline{){\displaystyle 230.67}}$

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