solve the below equation with respect to x

$0.6\cdot \mathrm{exp}(\frac{-40}{x})+0.4\cdot \mathrm{exp}(\frac{10}{x})=1$

$0.6\cdot \mathrm{exp}(\frac{-40}{x})+0.4\cdot \mathrm{exp}(\frac{10}{x})=1$

Cristofer Watson
2022-10-21
Answered

solve the below equation with respect to x

$0.6\cdot \mathrm{exp}(\frac{-40}{x})+0.4\cdot \mathrm{exp}(\frac{10}{x})=1$

$0.6\cdot \mathrm{exp}(\frac{-40}{x})+0.4\cdot \mathrm{exp}(\frac{10}{x})=1$

You can still ask an expert for help

Amadek6

Answered 2022-10-22
Author has **21** answers

Equations which are sums of exponential functions do not show analytical solutions in general.

Your case seems among the particular ones since writing $y={e}^{\frac{10}{x}}$ the euation writes

$$\frac{0.6}{{y}^{4}}+0.4y=1$$

which, unfortunately reduces to a quintic polynomial which is not the most pleasant to solve (few quintic equations have analytical solutions - if they have, they tend to show many radicals).

Probably the easiest could be to look for the zero of

$$f(y)=0.4{y}^{5}-{y}^{4}+0.6$$

If you graph the function, you could notice that the positive root is around 2.5. So, using Newton method for example

$${f}^{\prime}(y)=2{y}^{4}-4{y}^{3}$$

gives the iterative scheme

$${y}_{n+1}=\frac{8{y}_{n}^{5}-15{y}_{n}^{4}-3}{10({y}_{n}-2){y}_{n}^{3}}$$

Using ${y}_{0}=2.5$, the interates will then be

$${y}_{1}=2.46160000000000$$

$${y}_{2}=2.45898401374040$$

$${y}_{3}=2.45897234660929$$

$${y}_{4}=2.45897234637796$$

which is the solution for fifteen significant figures.

Now, from y get x.

Your case seems among the particular ones since writing $y={e}^{\frac{10}{x}}$ the euation writes

$$\frac{0.6}{{y}^{4}}+0.4y=1$$

which, unfortunately reduces to a quintic polynomial which is not the most pleasant to solve (few quintic equations have analytical solutions - if they have, they tend to show many radicals).

Probably the easiest could be to look for the zero of

$$f(y)=0.4{y}^{5}-{y}^{4}+0.6$$

If you graph the function, you could notice that the positive root is around 2.5. So, using Newton method for example

$${f}^{\prime}(y)=2{y}^{4}-4{y}^{3}$$

gives the iterative scheme

$${y}_{n+1}=\frac{8{y}_{n}^{5}-15{y}_{n}^{4}-3}{10({y}_{n}-2){y}_{n}^{3}}$$

Using ${y}_{0}=2.5$, the interates will then be

$${y}_{1}=2.46160000000000$$

$${y}_{2}=2.45898401374040$$

$${y}_{3}=2.45897234660929$$

$${y}_{4}=2.45897234637796$$

which is the solution for fifteen significant figures.

Now, from y get x.

Seettiffrourfk6

Answered 2022-10-23
Author has **1** answers

The solution of the equation can be expressed with radicals. Be ready for the monster

$$y=\frac{3}{8}+\frac{1}{8}\sqrt{25-\frac{16\text{}{5}^{2/3}}{\sqrt[3]{\sqrt{65}-5}}+8\sqrt[3]{5(\sqrt{65}-5)}}+\frac{1}{2}\sqrt{\frac{25}{8}+\frac{{5}^{2/3}}{\sqrt[3]{\sqrt{65}-5}}-\frac{1}{2}\sqrt[3]{5(\sqrt{65}-5)}+\frac{195}{8\sqrt{25-\frac{16\text{}{5}^{2/3}}{\sqrt[3]{\sqrt{65}-5}}+8\sqrt[3]{5(\sqrt{65}-5)}}}}$$

$$y=\frac{3}{8}+\frac{1}{8}\sqrt{25-\frac{16\text{}{5}^{2/3}}{\sqrt[3]{\sqrt{65}-5}}+8\sqrt[3]{5(\sqrt{65}-5)}}+\frac{1}{2}\sqrt{\frac{25}{8}+\frac{{5}^{2/3}}{\sqrt[3]{\sqrt{65}-5}}-\frac{1}{2}\sqrt[3]{5(\sqrt{65}-5)}+\frac{195}{8\sqrt{25-\frac{16\text{}{5}^{2/3}}{\sqrt[3]{\sqrt{65}-5}}+8\sqrt[3]{5(\sqrt{65}-5)}}}}$$

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