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I am trying to solve the following equation for $x$, a simplifying assumption I can make is that $N$ is very large.

$$1= \frac{N}{e^{10ax}+e^{9ax}+1}$$

$a$ is an arbitrary constant.

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    $\begingroup$ if $N$ is very large you can probably remove $+1$ from the denominator but I do not see how this can be solved for $x$ without numerical methods $\endgroup$ – Vasya Feb 21 at 18:55
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    $\begingroup$ This is equivalent to the 10th degree equation $u^{10}+u^9-(N+1)=0$, where $u=e^{ax}$. It will require numerical method to find the value of $u$. $\endgroup$ – Alain Remillard Feb 21 at 19:32
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Assume $N\gg 1$. Then, the given equation indicates that $ax\gg1$. Let $t = e^{-ax}\ll 1$ and rewrite the equation as,

$$t^{10}=\frac{1+t}{N-1}$$

or, in approximate form, $$ t=\left(\frac{1+t}{N-1}\right)^{1/10}=\frac1{{N}^{1/10}}\left(1+\frac t{10} + O(t^2)\right)$$

The leading order solution is then $t_0 = \frac1{{N}^{1/10}}$ and the solution with the first two terms is,

$$t = \frac1{{N}^{1/10}}\left(1+ \frac1{10{N}^{1/10}}\right)$$

Correspondingly, the solution for $x$ is,

$$x=-\frac1a \left[ \ln \frac1{{N}^{1/10}}+ \ln\left(1+\frac1{10{N}^{1/10}}\right)\right] =\frac1{10a} \left( \ln N -\frac 1{N^{1/10}}\right) $$

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Just for the fun of it !

Starting from @Quanto's answer$$t^{10}=\frac{1+t}{N-1}\implies t=\frac{1}{\sqrt[10]{N-1}}(1+t)^{\frac 1 {10}}$$ Then, we look for the zero of function $$f(t)=t-k(1+t)^{\frac 1 {10}}\qquad \text{with}\qquad k=(N-1)^{-\frac 1 {10}}$$ which can be solved using high order iteration methods starting with $t_0=k$.

This gives the result in the form $$t=k \left(1+\sum_{n=1}^p \frac{b_n}{c_n}k^n\right)$$

The first coefficients are given below $$\left( \begin{array}{ccc} n & b_n & c_n \\ 1 & 1 & 10 \\ 2 & -7 & 200 \\ 3 & 2 & 125 \\ 4 & -1 & 128 \\ 5 & 119 & 31250 \\ 6 & -141427 & 80000000 \\ 7 & 286 & 390625 \\ 8 & -14498297 & 64000000000 \\ 9 & 0 & 1 \\ 10 & 5375265623 & 64000000000000 \\ 11 & -121771 & 1220703125 \\ 12 & 2223135501027 & 25600000000000000 \\ 13 & -1991846 & 30517578125 \\ 14 & 7429 & 167772160 \\ 15 & -104475642 & 3814697265625 \\ 16 & 624864674543698631 & 40960000000000000000000 \\ 17 & -1392689353 & 190734863281250 \\ 18 & 20827196416176078951 & 8192000000000000000000000 \\ 19 & 0 & 1 \\ 20 & -18524404899326965878239 & 16384000000000000000000000000 \end{array} \right)$$

Applied to a small number $N=1025$ (meaning $k=\frac 12$), this gives $$t=0.521428989990757$$ while the exact solution is $$t=0.521428989990988$$ Applied to a large number $N=(10^{100}+1)$ (meaning $k=10^{-10}$), this gives ore than $200$ exact significant figures.

Continuing to $x$, this will give $$a x=-\log (k)+\sum_{n=1}^p \frac{d_n}{e_n}k^n$$ for which the first coefficients are reported below $$\left( \begin{array}{ccc} n & d_n & e_n \\ 1 & -1 & 10 \\ 2 & 1 & 25 \\ 3 & -119 & 6000 \\ 4 & 13 & 1250 \\ 5 & -7 & 1280 \\ 6 & 1309 & 468750 \\ 7 & -7495631 & 5600000000 \\ 8 & 4433 & 7812500 \\ 9 & -1029379087 & 5760000000000 \\ 10 & 0 & 1 \\ 11 & 478398640447 & 7040000000000000 \\ 12 & -5966779 & 73242187500 \\ 13 & 237875498609889 & 3328000000000000000 \\ 14 & -57763534 & 1068115234375 \\ 15 & 7429 & 201326592 \\ 16 & -3499934007 & 152587890625000 \\ 17 & 89355648459748904233 & 6963200000000000000000000 \\ 18 & -26461097707 & 4291534423828125 \\ 19 & 3353178623004348711111 & 1556480000000000000000000000 \\ 20 & 0 & 1 \end{array} \right)$$

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