Related to this, I am looking for a solution for:
$(\log_{x}{x})'$ = ?
...where $x$ is not 1, but positive.
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7$$\log_bb=1$$ if $b>0,\ne1$ – lab bhattacharjee Oct 21 '13 at 18:54
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@labbhattacharjee Oh, so simple and clean. Post an answer! :-) – Ionică Bizău Oct 21 '13 at 18:57
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@labbhattacharjee Asked this answer my math teacher, and he didn't know to answer... ^_^ – Ionică Bizău Oct 21 '13 at 18:57
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3@Johnツ, if that's true then change of teacher...or of school. – DonAntonio Oct 21 '13 at 19:13
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1@DonAntonio: Mistakes happen sometimes. Have you never a dumb moment? – Najib Idrissi Oct 22 '13 at 00:56
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I can't believe I stared at this for five minutes before seeing what was going on! I actually tried $y = \log_x x \Rightarrow x^y = x \Rightarrow y \ln x = \ln x \Rightarrow y = 1$ before I figured it out! Question of Great Good Humor! ;-) – Robert Lewis Oct 22 '13 at 01:05
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Oh, yes I have, @nik ! But like this one and from a teacher? Hmmm...even if (s)he didn't notice immediately that $;\log_xx=1;$ , she could have readily tried change of basis: $$\log_xx=\frac{\log_e x}{\log_e x}=1$$ That (s)he didn't come up with the answer makes me wonder... – DonAntonio Oct 22 '13 at 04:21
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@DonAntonio Maybe an answer would be that I asked this question after I asked this one. I have a good math teacher. (S)he was so happy to find the answer today. Thank you! :-) – Ionică Bizău Oct 22 '13 at 18:36
2 Answers
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Notice that $\log_x x=1.$ Is that enough?
Maxim Gilula
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Remember: $\log_b a$ is the number we raise $b$ to to get $a.$ In other words, $\log_b a= c$ if and only if $b^c=a.$ – Maxim Gilula Oct 22 '13 at 02:16
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If you want to do it the hard way, let $f(y,z) = \log_y z = \frac{\ln z}{\ln y}$, so that your function is $g(x) = f(x,x)$. It is easy to compute that $\frac{\partial f}{\partial z}(y,z) = \frac{1}{z \ln y}.$ It is almost as easy to compute that $\frac{\partial f}{\partial z}(y,z) = \frac{1}{y} \cdot \frac{-\ln z}{\ln^2 y}$.
You have: $$ g'(x) = \frac{\partial f}{\partial y}(x,x) + \frac{\partial f}{\partial z}(x,x).$$ When you substitute the above formulas, everything cancels out and you find $g'(x) =0$.
Yes, it is somewhat silly to solve this particular problem this way. But hopefully, this will be of use to the author of the question if he wants to differentiate, say, $\log_x(1+x)$.
Jakub Konieczny
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