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?$ $\endgroup$– lab bhattacharjeeAug 28, 2016 at 10:24
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3 Answers
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$$a^{(\log_ab)^2}=a^{\log_a(b)\times \log_a(b)}=b^{\log_a(b)}$$
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$\begingroup$ Although, I did not vote down, but are you sure this is correct? $\endgroup$– ً ًAug 28, 2016 at 10:52
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$\begingroup$ 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. $\endgroup$– EvaristeAug 28, 2016 at 10:53
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1$\begingroup$ 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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3$\begingroup$ 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 $\endgroup$ 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:
$$a^{\left(\log_a(b)\right)^2}=a^{\left(\frac{\ln(b)}{\ln(a)}\right)^2}=a^{\frac{\ln^2(b)}{\ln^2(a)}}=\exp\left[\frac{\ln^2(b)}{\ln(a)}\right]$$
answered Aug 28, 2016 at 10:35
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$$a^{(\log_ab)^2}=a^{\log_a b\cdot \log_a b}=(a^{\log_a b})^{\log_a b}=b^{\log_a b}=b^{\frac{1}{\log_b a}}.$$
