6

I posted an answer in this question to prove that $(\sqrt8)^{\sqrt 7}<(\sqrt7)^{\sqrt 8}$

I started with

$$f(x)=\frac{\ln x}{x}$$

$$f'(x)=\frac{1-\ln x}{x^2}$$

$$f'(x)>0 : x\in)0,e($$ $$f'(x)<0 : x\in)e,+\infty($$

so $f(1/7)>f(1/8)$, hence $-8\ln 8<-7\ln7$

so

$$\begin{align} 8\ln 8 & > 7\ln7\\\\ 0.5\cdot 8\ln 8 & >0.5\cdot 7\ln7\\\\ 8\ln \sqrt8 & > 7\ln\sqrt7\\\\ \ln (\sqrt8)^{\sqrt 7}& >\ln(\sqrt7)^{\sqrt 8}\\\\ (\sqrt8)^{\sqrt 7} &>(\sqrt7)^{\sqrt 8} \end{align}$$

but where is the mistake with my answer, because $(\sqrt8)^{\sqrt 7}<(\sqrt7)^{\sqrt 8}$?!

Community
  • 1
mnsh
  • 5,875
  • 3
    How did you go from the third to last step to the second to last step? I could see how you might arrive at $\sqrt(8)^{\frac{1}{7}} > \sqrt{7}^{\frac{1}{8}}$, which is true. But I don't see how your third to last step implies your second to last step. – Alex Wertheim Aug 08 '13 at 02:19
  • 1
    @hmedan.mnsh: Just by the way, it is not correct to say "Where is the wrong?". You should use a word such as "mistake" or "error" instead of "wrong". – Zev Chonoles Aug 08 '13 at 02:24
  • @AWertheim but I know that if $g'(x)> 0$ for every $x\in D$ so if a>b and $a,b \in D$ its mean g(a)>g(b), if $g(x) =\ln x$ so $\ln a >\ln b$ for a>b – mnsh Aug 08 '13 at 02:27
  • @ZevChonoles thanks and at the next time I will do that ^_^ – mnsh Aug 08 '13 at 02:28
  • 3
    @hmedan.mnsh, while that's true I'm afraid I don't see how that's relevant. How precisely did you propose to go from $8\ln(\sqrt{8}) > 7 \ln(\sqrt{7})$ to $\ln(\sqrt{8}^{\sqrt{7}}) > \ln(\sqrt{7}^{\sqrt{8}})$ Is it possible that you are confusing $x^{\frac{1}{7}}$ with $x^{\sqrt{7}}$? – Alex Wertheim Aug 08 '13 at 02:30
  • @AWertheim hehehehe because of sleepless but thanks alot – mnsh Aug 08 '13 at 02:35
  • 1
    No problem, glad we could get that straightened out! :) – Alex Wertheim Aug 08 '13 at 02:36

1 Answers1

6

In your answer, you made the following comment:

@BarryCipra $8\ln\sqrt8>7\ln\sqrt7$, so $1/7∗1/8∗8\ln\sqrt8>1/7∗1/8∗7\ln\sqrt7$, so $1/7\ln\sqrt8>1/8\ln\sqrt7$ and note that $a\ln b=\ln b^a$.

It is true that: $$ \dfrac{1}{7}\ln\sqrt8 > \dfrac{1}{8}\ln\sqrt7 $$

However, if you apply the power rule for logs, you instead obtain: $$ \ln\sqrt[7]{\sqrt8} > \ln\sqrt[8]{\sqrt7} $$ which is true, but not quite what we wanted.

Adriano
  • 41,576