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Can the following expression be further simplified: $$a^{(\log_ab)^2}?$$

I know for example that $$a^{\log_ab^2}=b^2.$$

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  • $\begingroup$ $$a^{mn}=(a^m)^n$$ $ m=n\implies?$ $\endgroup$ Aug 28, 2016 at 10:24

3 Answers 3

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$$a^{(\log_ab)^2}=a^{\log_a(b)\times \log_a(b)}=b^{\log_a(b)}$$

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  • $\begingroup$ What's so wrong with this ?... $\endgroup$
    – Evariste
    Aug 28, 2016 at 10:46
  • $\begingroup$ Although, I did not vote down, but are you sure this is correct? $\endgroup$
    – ً ً
    Aug 28, 2016 at 10:52
  • $\begingroup$ This probably isn't, but I'd like an explanation before I remove it for good, just so I don't make the same mistake in the future. I'm genuinely confused. $\endgroup$
    – Evariste
    Aug 28, 2016 at 10:53
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    $\begingroup$ If $\log_a b$ is a real number (as we can suppose from OP) it seems correct to me. $\endgroup$ Aug 28, 2016 at 10:56
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    $\begingroup$ Unless otherwise explicitly stated, these questions should, imo, be considered as asking about the real, standard logarithm, and thus this answer is correct. Of course, it must be $\;a,b>0\,,\,\,a\neq1\;$ and etc. Whoever downvoted was trolling around (there are quite a few around...), or was just bored, or just couldn't understand this answer. +1 $\endgroup$
    – DonAntonio
    Aug 28, 2016 at 11:05
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Use:

  • $$\log_x(y)=\frac{\ln(y)}{\ln(x)}$$
  • $$\exp\left[\ln\left(x\right)\right]=e^{\ln(x)}=x$$

So, we get:

$$a^{\left(\log_a(b)\right)^2}=a^{\left(\frac{\ln(b)}{\ln(a)}\right)^2}=a^{\frac{\ln^2(b)}{\ln^2(a)}}=\exp\left[\frac{\ln^2(b)}{\ln(a)}\right]$$

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$$a^{(\log_ab)^2}=a^{\log_a b\cdot \log_a b}=(a^{\log_a b})^{\log_a b}=b^{\log_a b}=b^{\frac{1}{\log_b a}}.$$

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  • $\begingroup$ Is $\log_{a} {b} = -\log _{b}{a}$ ? $\endgroup$
    – Airdish
    Aug 28, 2016 at 10:39

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