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So the ln rule applies to $\ln(a^b)$ by converting it into $b\ln(a)$, but what about if there's a power applied to the $b$? Such as $\ln(a^{b^c})$. I'm sorry I can't format this.

Does it become $b^c\ln(a)$ or $c(b\ln(a))$, where the higher power is moved into the front? I'm thinking it's the former, but I couldn't find this on the internet so any help would be great.

Henry
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Mathhurtsmybrain
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    $\ln(a^x) = x \ln(a)$, so when $x = b^c$ we then have $\ln(a^{b^c}) = b^c \ln(a)$. Note also that $\ln(\ln(a^{b^c})) = c \ln(b) + \ln(a)$ using the fact that the $\ln$ of a product is the sum of the $\ln$'s. – azif00 Sep 02 '21 at 00:58
  • $(\ln e)^a^b$ is ambiguous. Do you mean $(\ln e)^{(a^b)}$ or $((\ln e)^a)^b$? – Steven Alexis Gregory Sep 02 '21 at 01:02
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    It is usual to read $a^{b^c}$ as $a^{(b^c)}$ leading to the first of your suggestions $\ln(a^{b^c}) = b^c \ln(a)$. Meanwhile $\ln((a^b)^c) = c\ln(a^b) = cb\ln(a)$ though you can get there more quickly using $(a^b)^c=a^{bc}$ – Henry Sep 02 '21 at 01:03

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The answer would be $b^c\ln(a)$ to your question that is $\ln(a^{b^c})$

If it were $\ln((a^b)^c)$ then the answer would be $bc\ln(a)$.

Ilovemath
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