Is $\log (e^n+n^e) =n$ for $n\geq15$ or for large $n$?

Question

I have tried some calculations for this formula $\log (e^n+n^e) $ i found that is closed to integer from $n=15$ which is $n$ , or probably to conjecture that $\log (e^n+n^e) $ could be integer and close to $n$ from $n=15$.

My question here is: Is $\log (e^n+n^e) =n$ for $n\geq15$ with $\log$ is natural logarithm?

Answer 1

Note that $$\log(e^n+n^e) > \log(e^n)=n.$$

Answer 2

Note that since $\log$ function in strictly monotonic increasing

$$x>y \iff \log x > \log y$$

and since

$$e^n+n^e>e^n \implies \log (e^n+n^e)>\log e^n=n$$

$$\implies \log (e^n+n^e)>n$$

What is true is that for $n\to\infty$

$$\log (e^n+n^e)\sim n$$

notably we say that for n large $\log (e^n+n^e)$ is asymptotically equivalent to $n$ in the sense that $$\lim_{n\to\infty} \frac{\log (e^n+n^e)}{n}=1$$

Answer 3

No since that would imply that $$ e^n+n^e=e^n\implies n^e=0 $$ a contradiction.

Answer 4

Since $e^n \lt e^n + n^e \lt e^{n+1}$ for $n > 0$ (proof left as an exercise), $n \lt \ln (e^n + n^e) \lt n + 1$, so the logarithm will never equal an integer (because it's always stuck between two of them). For sufficiently large $n$, it will get arbitrarily close to equalling $n$. So it will never be exactly equal, but for any positive value $\epsilon > 0$ you can make the difference smaller than $\epsilon$.

Answer 5

By Taylor, $$\ln(e^n+n^e)=n+\ln(1+n^ee^{-n})=n+n^ee^{-n}-\frac12(n^ee^{-n})^2\pm\dots.$$ For large $n$, $n^ee^{-n}$ and thus the fractional part of the expression will be small and rapidly falling.

For double floating point accuracy the fractional term will be too small to influence the mantissa if the relative error is smaller than the machine epsilon, $$ n^{e-1}e^{-n}\le 2^{-53}\iff -\frac{n}{e-1}e^{-\frac{n}{e-1}}\ge-\frac{2^{-53/(e-1)}}{e-1} $$ which with the Lambert-W function gives $$ n\ge -(e-1)W_{-1}\left(-\frac{2^{-53/(e-1)}}{e-1}\right)=43.2078896438... $$ Thus indeed for $n\ge44$ the numerical result in double precision floating point will be $n$.

 n    ln(e^n+n^e)

37   37.0000000000015631940186722
38   38.0000000000006181721801113
39   39.0000000000002415845301584
40   40.0000000000000994759830064
41   41.0000000000000355271367880
42   42.0000000000000142108547152
43   43.0000000000000071054273576
44   44.0000000000000000000000000