Function $f(x)=(\ln x)^n-x$ If $f(x)=(\ln x)^n-x$ then
(A*) $f(x) = 0$ has exactly two solutions for $n = 5$
(B) $f(x) = 0$ has exactly one solution for $n = 5$
(C*) $f(x) = 0$ has no solutions if $n\in(0,1)$
(D) $f(x) = 0$ has exactly two solutions if $n\in(0,1)$
My approach is as follow
The correct answer is A and C but as per my approach I am getting $f(x)=(\ln x)^n-x=0$ so $(\ln x)^n=x$
I checked the answer on desmos.com when n=5 it intersect at one point only . Hence not able to prove the answer
 A: if $n$ is an odd integer, the function $f$ is defined on $(0,\infty)$ and tends to $-\infty$ for $x\to+\infty$ as well as for $x\to0^+$.
To count the zeroes, we can consider $g(x)=f(e^{nx})=(nx)^n-e^{nx}$ on $\Bbb R$ instead.
In particular, for $n=5$ we have $g(1)=f(e^5)=5^5-e^5>0$, so by continuity have at least one zero of $g$ in $(-\infty,1)$ and at least one zero in $(1,\infty)$.
Note that $g(x)=0$ iff $(nx)^n=e^{nx}$, iff $nx=e^{x}$. Now $x\mapsto e^x$ is well-known to be convex, i.e., can intersect a line such as $x\mapsto nx$ in no more than two points. We conclude that (A*) is true and (B) is false.
If $0<n<1$, then $f$ is defined only for $x\ge 1$.
Note that $y^n\le\max\{1,x\}$ for $y\ge0$.
Hence with $y:=\ln x\ge0$ and using the well-known inequality $e^y\ge 1+y$, we have
$f(x)=y^n-e^y \stackrel {(1)}\le\max\{1,y\}-e^y\stackrel {(2)}\le 1+y-e^y\stackrel {(3)}\le 0$
with equality at $(1)$ only when $y=1$, at $(2)$ and $(3)$ only when $y=0$, hence overall never. We conclude that (C*) is correct and (D) is false.
