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I'm not quite sure why this is true, perhaps a better explanation will help me out. I understand that a natural log has to be a positive number.

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5 Answers

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$e = e^1$

$1 = e^0$

So $\ln(e) =1$ and $\ln(1) = 0$.

And $\ln(\ln e) = \ln (1) = 0$.

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$\ln e = 1$.

You just have to evaluate $\ln 1$.

The input of a natural log has to be positive, the output need not be positive.

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$\ln x:= n$ such that $e^n=x$. So, $\ln e=n$ such that $e^n=e.$ So, $\ln e=1$. Now, $\ln 1=n$ such that $e^n=1$, which implies that $\ln 1=0$.

$$ \ln(\ln e)=\ln(1)=0.$$

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The natural log of a number is whatever you have to raise $e$ to, in order to get that number. So for example $\ln(10)$ is saying, "what do I need to raise $e$ to, in order to get 10?. So you could plug that in your calculator, and see $\ln(10) \approx 2.3025$ and then you could check it by seeing what $e^{2.3025}$ is.

In your case, $\ln(e)$ is saying, "what do I need to raise $e$ to, in order to get $e$? well obviously that's just 1 since $e^1 = e$. So then we are left with $\ln(1)$. "What do I need to raise $e$ to in order to get 1?". Well raising anything to the 0 power gives you 1. So the answer of what you need to raise $e$ to is, zero.

Thus $\ln(\ln(e)) = 0$

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$$\ln(\ln e)=\ln 1= 0$$ The logarithm on base $a$ of $a$ is always $1$, since $a^{1} = a$. In addition, it has to be $a> 0$ and $a$ other than $1$, so that there are no problems.

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