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If we write the equation as $ax= (1+1/x)^{k-1}$ it is clear that there is exactly one positive solution (here and below I assume that $k$ is a real number larger than 1).

I. With the substitution $u=1/x$ the equation writes $a=u(1+u)^{k-1}$, and the RHS has a formal series inverse at $0$. The Lagrange inversion formula provides the (Laurent) expansion at $0$ of any integer power of the latter, in particular, of its reciprocal. One finds
$$x=a^{-1}+(k-1)+ \sum_{n=1}^\infty {(-1)^n \over n} {nk \choose n+1} a^n\ ,$$ according with the first terms found by Robert Israel. This series has a radius of convergence $k^{-k}(k-1)^{k-1}$, so it gives the positive solution of the initial equation provided $a < k ^ {-k}(k-1) ^ {k-1}$. Moreover the terms of the series are eventually decreasing in absolute value, which implies that the $n$-th partial sums are eventually a bound on $x$ alternately from above and from below. If needed, one may also estimate the remainder from the partial sums using bounds on the binomial coefficients.

II.  For large positive values of $a$, we can write the equation in the form $$a^{-\frac{1}{k}}=x(1+x)^{-1+\frac{1}{k} } \ .$$ Again, the Lagrange inversion formula provides a power series expansion of the inverse of the RHS; computing the radius of convergence this time we find $k (k-1)^{\frac{1}{k}-1 }$. Hence, evaluating that inverse at $a^{-\frac{1}{k}}$ we get $$x=\sum_{n=1}^\infty {1\over n}{\frac{n(k-1)}{k} \choose n-1 }a^{-{n\over k}}\ ,$$ converging for $a>k^{-k}(k-1)^{k-1}$. Again, estimates on the convergence are available with a little more work on the binomial coefficients. For $a=1/2$ and $k=5$, the first $20$ terms of the series sum to $x=4.4786$; compare with the numerical solution in Aleksey Pichugin's answer below.)

III. It remains the case $a := k^{-k} (k-1)^{k-1}$, though it should be covered as a limit case by the above series.

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If we write the equation as $ax= (1+1/x)^{k-1}$ it is clear that there is exactly one positive solution (here and below I assume that $k$ is a real number larger than 1).

I. With the substitution $u=1/x$ the equation writes $a=u(1+u)^{k-1}$, and the RHS has a formal series inverse at $0$. The Lagrange inversion formula provides the (Laurent) expansion at $0$ of any integer power of the latter, in particular, of its reciprocal. One finds
$$x=a^{-1}+(k-1)+ \sum_{n=1}^\infty {(-1)^n \over n} {nk \choose n+1} a^n\ ,$$ according with the first terms found by Robert Israel. This series has a radius of convergence $k^{-k}(k-1)^{k-1}$, so it gives the positive solution of the initial equation provided $a < k ^ {-k}(k-1) ^ {k-1}$. Moreover the terms of the series are eventually decreasing in absolute value, which implies that the $n$-th partial sums are eventually a bound on $x$ alternately from above and from below. If needed, one may also estimate the remainder from the partial sums using bounds on the binomial coefficients.

II.  For large positive values of $a$, one we can write the equation in the form $$a^{-\frac{1}{k}}=x(1+x)^{-1+\frac{1}{k} } \ .$$ Again, the Lagrange inversion formula provides a power series expansion of the RHS; computing the radius of convergence this time one finds we find $k (k-1)^{\frac{1}{k}-1 }$. Hence, evaluating the that inverse at $a^{-\frac{1}{k}}$ we get $$x=\sum_{n=1}^\infty {1\over n}{\frac{n(k-1)}{k} \choose n-1 }a^{-{n\over k}}\ ,$$ converging for $a>k^{-k}(k-1)^{k-1}$. Again, estimates on the convergence are available with a little more work on the binomial coefficients; anyway now the convergence is quite fast, due to the decay of the binomial coefficients. For $a=1/2$ and $k=5$, the first $20$ terms of the series sum to $x=4.4786$; compare with the numerical solution in Aleksey Pichugin's answer below.)

III. It remains therefore the case $a _ 0 := k^{-k} (k-1)^{k-1}$; although , though it should be covered as a limit case by the above series.

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If we write the equation as $ax= (1+1/x)^{k-1}$ it is clear that there is exactly one positive solution (here and below I assume that $k$ is a real number larger than 1).

I. With the substitution $u=1/x$ the equation writes $a=u(1+u)^{k-1}$, and the RHS has a formal series inverse at $0$. The Lagrange inversion formula provides the (Laurent) expansion at $0$ of any integer power of the latter, in particular, of its reciprocal. One finds
$$x=a^{-1}+(k-1)+ \sum_{n=1}^\infty {(-1)^n \over n} {nk \choose n+1} a^n\ ,$$ according with the first terms found by Robert Israel. This series has a radius of convergence $k^{-k}(k-1)^{k-1}$, so it gives the positive solution of the initial equation provided $a II.  For large positive values of$a$, one can write the equation in the form $$a^{-\frac{1}{k}}=x(1+x)^{-1+\frac{1}{k} } \ .$$ Again, the Lagrange inversion formula provides a power series expansion of the RHS; computing the radius of convergence this time one finds$k (k-1)^{\frac{1}{k}-1 }$. Hence, evaluating the inverse at$a^{-\frac{1}{k}}$$$x=\sum_{k=1}^\infty x=\sum_{n=1}^\infty {1\over n}{\frac{n(k-1)}{k} \choose n-1 }a^{-{n\over k}}\ ,$$ converging for$a>k^{-k}(k-1)^{k-1}$. Again, estimates on the convergence are available with a little more work on the binomial coefficients; anyway now the convergence is quite fast, due to the decay of the binomial coefficients. For$a=1/2$and$k=5$, the first$20$terms of the series sum to$x=4.4786$; compare with the numerical solution in Aleksey Pichugin's answer below.) III. It remains therefore the case$ a _ 0 := k^{-k} (k-1)^{k-1} $; although it should be covered as a limit case by the above series, it could be nice to see a power series solution for values of$a$in a neighbourhood of$a_0\$..

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