# solve the below equation with respect to x 0.6*exp(−40/x)+0.4*exp(10/x)=1

solve the below equation with respect to x
$0.6\cdot \mathrm{exp}\left(\frac{-40}{x}\right)+0.4\cdot \mathrm{exp}\left(\frac{10}{x}\right)=1$
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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
$f\left(y\right)=0.4{y}^{5}-{y}^{4}+0.6$
If you graph the function, you could notice that the positive root is around 2.5. So, using Newton method for example
${f}^{\prime }\left(y\right)=2{y}^{4}-4{y}^{3}$
gives the iterative scheme
${y}_{n+1}=\frac{8{y}_{n}^{5}-15{y}_{n}^{4}-3}{10\left({y}_{n}-2\right){y}_{n}^{3}}$
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.
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Seettiffrourfk6
The solution of the equation can be expressed with radicals. Be ready for the monster