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As I know following statements are correct:

$a^{\log_bc} = (b^{\log_ba})^{\log_bc} = b^{\log_ba\log_bc} = (b^{\log_bc})^{\log_ba} = c^{\log_ba},$

$\log_aa^x = x,$

$\log_ab^c = c\log_ab.$

Then I don't get where is the mistake below:

$ a^{\log_bc} = (\log_mm^a)^{\log_bc} = \log_mm^{a\log_bc} = a\log_bc$

or alternatively

$ a^{\log_bc} = (\log_mm^a)^{\log_bc} = \log_mm^{a\log_bc} = (\log_mm^{\log_bc})^a = \log_bc^a = a\log_bc$

Obviously there is a mistake(s) in above reasoning, but I don't get it.

amWhy
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mathdummy
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    I don't understand $(\log_mm^a)^{\log_bc} = \log_mm^{a\log_bc}$. – Randall Feb 08 '23 at 14:52
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    In cases like this, where you "know" that there's a computational error somewhere but you can't spot it, it can be useful to pick particular values. then go through the computation step by step using those numbers until you come to a false equality. – lulu Feb 08 '23 at 14:59
  • So, is it wrong? Can't do that? Why? – mathdummy Feb 08 '23 at 14:59
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    $m=2$, $a=3$, $b=2$, $c=16$. Compute both sides. One side is 81, the other is 12. – Randall Feb 08 '23 at 15:03
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    @mathdummy using parentheses might help: $\color{red}{(\log_m(m^a))^{\log_bc} \neq (\log_m(m))^{a\log_bc}}$. – D S Feb 08 '23 at 15:31
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    Note that $(\log x^2)^2 \ne \log x^4$ – Vasili Feb 08 '23 at 15:56

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