Question

Solve for n:
log n + log n log 2 <= n

Answer 1

Is this base 10 or base e?
And, is this the case?
\[
\log(n)+\log(n)\cdot\log(2)=n
\]

Answer 2

we can assume it's base e

Answer 3

Oh, wait, that was a stupid question, never mind, here:
\[
\log(n)+\log(n)\cdot\log(2)\leq n
\]
There's no elementary way to solve this, just a note (not that I can see, at least). But, we know that:
\[
\frac{d}{dn}\log(n)(1+\log(2))\le \frac{d}{dn}n, n\geq1+\log(2)
\]And (with some quick algebra):
\[
\log(n)(1+\log(2))\lt n, n\in (1, 1+\log(2))
\]And it is obvious to see that:
\[
\log(n)(1+\log(2))\lt n, n\in (0, 1]
\]Thus, we are done, and we see that:
\[
n\in(0, 1]\cup(1, 1+\log(2))\cup[1+\log(2), \infty) = (0, \infty)
\]Or:
\[
n\gt 0
\]