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So I was thinking about random maths when I couldn’t sleep last night, and I had an idea.

You can simplify $\ln{(a^b)}$ into $b\ln{(a)}$, and $\ln{(ab)}$ into $\ln{(a)}+\ln{(b)}$.

Observing this, it seems that $\ln$ turns power into multiplication, and multiplication can into addition, and by the sequence, $\ln{(^ba)}$ can be simplified into $\ln^b{(a)}$?

Is my statement true, and why / why not?

YesSpoon3
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  • Try $\ln(^2a)=\ln(a^a)$. You will see that $\ln(a^a)=a\ln a$ but your formula claims that $\ln(a^a)=\ln^2a$, which is not the same as $a\ln a$. – Aiden Chow Feb 22 '23 at 08:51

3 Answers3

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$\log_e\left(e^{e^e}\right)= e^e\log_e(e)=e^{e} \not = 1 =\left(\log_e(e)\right)^3$

and more generally $\log{(^ba)} = {^{b-1}a}\,\log(a)$ is usually not $\left(\log(a)\right)^b=\log^b(a)$

Henry
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No.

Counterexample: $\ln(^32) = \ln(2^4) = 4\ln 2 \neq (\ln 2)^3$.

More generally, the good formula is $\ln (^ba) = \ln (a^{(^{b-1}a)}) = (^{b-1}\!a) \ln a$.

LL 3.14
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No, for example: $$ \mathrm{ln}(^3e) = e^e\mathrm{ln}(e) = e^e \neq \mathrm{ln}^3(e) = 1. $$

Carlos Santi Toledo
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