# What is $a^{(\log_ab)^2}$?

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

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

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

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

• What's so wrong with this ?... Aug 28, 2016 at 10:46
• Although, I did not vote down, but are you sure this is correct? Aug 28, 2016 at 10:52
• 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. Aug 28, 2016 at 10:53
• If $\log_a b$ is a real number (as we can suppose from OP) it seems correct to me. Aug 28, 2016 at 10:56
• 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 Aug 28, 2016 at 11:05

Use:

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

So, we get:

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

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

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