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

Question

I have a question that I am not sure how to answer:

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

Answer 1

Because we need to disprove this statement, we only need to find 1 counterexample. $$(\log e)^2\neq 2(\log e)$$ Statement proved.

Answer 2

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 $k \ge 1$.