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I know that $$\ln e^2=2$$ But what about this? $$(\ln e)^2$$ A calculator gave 1. I'm really confused.

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    $\begingroup$ Note the following things: $\ln(x^2)=2\ln(x)$, $\ln(e)=1$ and $1^2=1$ $\endgroup$ – JMoravitz Dec 6 '17 at 3:56
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Consider the equality (assuming the operations are actually defined for m and n):

$$ x =\log _nm$$

What this means is that x is the number to you need to raise n to the power of, to get m. In other words:

$$ n^x = m $$

You probably already know this since your question stated:

$$\ln e^2=2$$ and the power you need to raise e to, to get e2, is two.


In the case where n and m are the same number, the logarithm will always be one:

$$ x^1 = x, \space \log_xx = 1$$ $$ e^1 = e, \space \log_ee = \ln e = 1$$

And, of course, the reason why you're getting one can be explained with:

$$ (\ln e)^2 = (1)^2 = 1 $$

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Since $\ln e = 1 $ So $(\ln e)^2 = 1$

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Consider the most important property of logarithm

$$ \log(m)^n = n \log (m) $$

So that, $$ \log_e(e)^2 = 2 \log_e(e) = 2 \ln(e) = 2 $$

And since, $$ [ln(e)^2] ≠ [ln(e)]^2 $$

As you might be thinking that $ ln(e)^2 $ is same as $ [ln(e)]^2 $ but that's not true.

Actually, $$ ln(e)^2= ln(e^2) $$

We have $$ [ln(e)]^2 = [1]^2 = 1 $$

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    $\begingroup$ Typesetting note: $\ln$ is more common to see than $ln$ and use \cdot or \times for multiplication (generally \cdot for multiplication of reals for anything about middle-school math and \times for very elementary arithmetic, cross products, or cartesian products etc...). For more typesetting tips, visit this page. $\endgroup$ – JMoravitz Dec 6 '17 at 5:07
  • $\begingroup$ @JMoravitz Done both things! $\endgroup$ – Ravi Prakash Dec 6 '17 at 5:58
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$\ln e$ is 1. So $(\ln e ) ^ 2 $ is one.

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The natural log are the log with base $e$ (euler's number our napier constant). Therefore $$ \ln (x) = \log_e(x)$$ When you put $x=e$, we have $\ln(e)$, but that is simply $1$. Therefore $\big(\ln(e)\big)^2=1$.

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