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As part of solving a problem, I am requires to solve this equation: $$x=n\ln \left (1 + \frac{x}{n} \right) - \ln \left (1 + \frac{x}{n} \right) $$

I know that if $x=0$, this solves the equation. My question is there any other solutions and how to find them.

Note: the $x$ can depend on $n$ (otherwise I doubt there would be any solutions besides $x=0$).

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  • $\begingroup$ I tried to put it in an exponential form, giving me $e^x = (n-1)ln(1 + \frac{x}{n}$ but this doesn't seem to help me a lot. My next guess would be to study the derivative. $\endgroup$ – user401855 Feb 23 '17 at 9:59
  • $\begingroup$ You can solve this with the Lambert W-function for $n\neq 1$. $\endgroup$ – user90369 Feb 23 '17 at 10:23
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    $\begingroup$ @user90369 That function is way beyond my level. $\endgroup$ – user401855 Feb 23 '17 at 10:26
  • $\begingroup$ Change your equation into the form $y=ze^z$, that's all. Also useful is the form $y=z^{1/z}$ because here it's easy to discuss the quantity of solutions. $\endgroup$ – user90369 Feb 23 '17 at 10:32
  • $\begingroup$ Define the function $f(x):=(n-1)\ln\left(1+\frac{x}{n}\right)-x$ and study where it zeros are, approximately, with the help of it derivative. Then you can approximate the zeros using the Newton's method. $\endgroup$ – Masacroso Feb 23 '17 at 11:27
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Let's start with a change of variables: let $ u = \frac{x}{n} $. Then $$ nu = n\ln{(1+u)}-\ln{(1+u)} = (n-1)\ln{(1+u)} $$ so $ \ln{(1+u)} = \frac{n-1}{n}u $. Exponentiating, $$ 1+u= e^{\frac{n-1}{n}u} $$ Again change variables to $ v = u + 1 $; then $$ v = e^{\frac{n-1}{n}(v-1)} $$ i.e. $$ e^{\frac{n-1}{n}}v = e^{\frac{n-1}{n}v} $$

Now, if $ n = 1 $, then $ v = 1 $, $ u = 0 $, and so $ x = 0 $.

So let's assume $ n \neq 1 $. Moving things from side to side and multiplying by $ \frac{1-n}{n} $, $$ \left(\frac{1-n}{n}\right)ve^{\frac{1-n}{n}v} = \left(\frac{1-n}{n}\right)e^{\frac{1-n}{n}} $$ One final change of variables is left: let $ w = \left(\frac{1-n}{n}\right)v $. Then $$ we^w = \left(\frac{1-n}{n}\right)e^{\frac{1-n}{n}} = \left(\frac{1}{n}-1\right)e^{\frac{1}{n}-1} $$ Clearly, $ w = \frac{1}{n} - 1 $. Now let's return to our original problem. Combining all our changes of variables, $$ x = nu = n(v-1) = \frac{n^2}{1-n}w = \frac{n^2}{1-n}\left(\frac{1}{n} - 1\right) = \frac{n^2}{n-1}\cdot\frac{1-n}{n} = -n $$

EDIT: You are right, technically, this doesn't solve your problem; setting $ x = -n $ yields $ \log{(0)} $ in the original equation. One could argue that instead, it suffices to solve the continuous extension, which is what we did. But we can do better. For n > 1, we have two solutions. To get the other one, just use the other branch of the Lambert W function: $$ x = nu = n(v-1) = \frac{n^2}{1-n}w = \frac{n^2}{1-n}W_{-1}\left(\left(\frac{1}{n}-1\right)e^{\frac{1}{n}-1}\right) $$

You said $ W $ is way above your level, so in a second, I'll have some asymptotics to think about. But ultimately this is not something that can be put in closed form.

EDIT 2:

As $ n \to 1^+ $, $ w \to -\infty $, so $ x \to \infty $.

As $ n \to \infty $, we can see $ w \to -\frac{1}{e} $, so $ x \to \infty $ again.

So your solution should be roughly "U-shaped." Most symbolic calculators offer the $ W $ (on Mathematica, $W_{-1} $ is ProductLog[-1, #]&), so you ought to be able to plot the result. Mathematica puts a minimum at $n=2.75$ with $x=6.4$; the plot seems to indicate that such is the only minimum. I don't think looking at series along the boundaries of your domain is useful either; you just get nasty messes at best. (I had to pull out the NIST Handbook of Mathematical Functions just to get started.)

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HINT: Begin by writing \begin{eqnarray*} x &=& \ln \left( 1 + \frac{x}{n} \right)^n - \ln \left( 1 +\frac{x}{n} \right) \\ \therefore x &=& \ln \left( 1 + \frac{x}{n} \right)^{n-1} \\ \end{eqnarray*}

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let's look at your first intuition about how to solve the problem. To do this we need to use the power rule and the quotient rule.

$$\begin {array} {ccc} & x&=&n*\ln(1+\frac{x}{n})-\ln(1+\frac{x}{n}) \\ & &=& \ln((1+\frac{x}{n})^n)-\ln(1+\frac{x}{n}) \\ & &=& \ln \left(\frac{(1+\frac{x}{n})^n}{1+\frac{x}{n}}\right) \\ & &=& \ln ((1+\frac{x}{n})^{n-1}) \\ \Rightarrow & e^x&=& (1+\frac{x}{n})^{n-1} \end {array}$$

From here lets define $f_n(x)= (1+\frac{x}{n})^{n}$. Now take the derivative. $f^\prime_n(x)= (1+\frac{x}{n})^{n-1}=e^x$. this would mean that $f_n(x)=\int e^x dx= e^x+c$.

This doesn't exactly solve it but it does simplify the solution.

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As I told in my comment above the form $\enspace\displaystyle y=z^{\frac{1}{z}}\enspace$ makes it easy to discuss the quantity of the solutions.

Your equation is with $\enspace n\neq 1\enspace$ and $\enspace z>0\enspace$ equivalent to $\enspace\displaystyle y=z^{\frac{1}{z}}\enspace$ with $\enspace\displaystyle y:=e^{\frac{n}{n-1} e^{-\frac{n}{n-1}}}$

and $\enspace\displaystyle z:= e^{\frac{n}{n-1}}\left(1+\frac{x}{n}\right)\enspace$ so that you get $\enspace x\enspace$ by $\enspace z\enspace$ (and $\enspace n\enspace$).

A discussion of the function $\enspace\displaystyle y=z^{\frac{1}{z}\enspace}$ shows:

$(1)\enspace$ $0<y\leq 1$ : $\enspace$ one solution

$(2)\enspace$ $\displaystyle 1<y<e^{\frac{1}{e}}$ : $\enspace$ two solutions

$(3)\enspace$ $\displaystyle y=e^{\frac{1}{e}}$ : $\enspace$ one solution

$(4)\enspace$ $\displaystyle y>e^{\frac{1}{e}}$ : $\enspace$ no solution

$\displaystyle y=z^{\frac{1}{z}}\enspace$ can be changed to $\enspace\displaystyle (-z\ln y)e^{-z\ln y}=-\ln y\enspace$ which can be solved

with the Lambert W function (with the upper and lower branch) by it's definition;

$\displaystyle z=\frac{W(-\ln y)}{-\ln y}\enspace$ and for a second solution $\enspace\displaystyle z=\frac{W_{-1}(-\ln y)}{-\ln y}$ .

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