I know that $$\ln e^2=2$$ But what about this? $$(\ln e)^2$$ A calculator gave 1. I'm really confused.
Consider the equality (assuming the operations are actually defined for
$$ x =\log _nm$$
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:
and the power you need to raise
e to, to get
e2, is two.
In the case where
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:
$$ (\ln e)^2 = (1)^2 = 1 $$
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 $$ [ln(e)]^2 = ^2 = 1 $$