<s>[INST] Show that $(\ln a)^k \neq k \ln a $ I have a question that I am not sure how to answer:

Show that $(\ln a)^k \neq k \ln a $

 [/INST] I will show that
if $k > 1$
then $a \le e$.
In other words,
if $a > e$
there is no $k \ge 1$
that satisfies this.
(This is just playing around
with algebra and
elementary calculus.)
If
$(\log a)^k=k\log a$,
then
$(\log a)^{k-1}=k$
or
$\log a
=k^{1/(k-1)}
$.
Therefore,
for any given $k$,
there is at most one $a$.
Let
$f(k) = k^{1/(k-1)}$
and
$g(k) = \ln f(k) =\frac{\ln k}{k-1}$.
$g'(k)
=\frac{(k-1)/k-(\ln k)}{(k-1)^2}
=\frac{1-1/k-\ln k}{(k-1)^2}
$.
For small $c$,
$g'(1+c)
=\frac{1-1/(1+c)-\ln (1+c)}{c^2}
\approx \frac{1-(1-c+c^2)-(c-c^2/2)}{c^2}
= \frac{(c-c^2)-c+c^2/2}{c^2}
= \frac{(c-c^2)-c+c^2/2}{c^2}
=-\frac12
$.
Similarly,
for small $c$,
$g(1+c)
=\frac{\ln (1+c)}{c}
\approx \frac{c-c^2/2}{c}
=1-c/2
$
so $g(1) = 1$
and
$f(1) = e$.
If $h(k)
=1-1/k-\ln k
$,
the numerator of $g(k)$,
then
$h'(k)
=1/k^2-1/k
=\frac{1-k}{k^2}
< 0
$
for $k > 1$.
Therefore,
$g$, and therefore $f$
is decreasing for $k \ge 1$,
so $a$ can be at most
$e$ for
$k \ge 1$.
</s>