I know that $$\ln e^2=2$$ But what about this? $$(\ln e)^2$$ A calculator gave 1. I'm really confused.
$\endgroup$ 15 Answers
$\begingroup$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:
$$ (\ln e)^2 = (1)^2 = 1 $$
$\endgroup$ $\begingroup$Since $\ln e = 1 $ So $(\ln e)^2 = 1$
$\endgroup$ $\begingroup$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 $$ [ln(e)]^2 = [1]^2 = 1 $$
$\endgroup$ 2 $\begingroup$$\ln e$ is 1. So $(\ln e ) ^ 2 $ is one.
$\endgroup$ $\begingroup$The natural log are the log with base $e$ (euler's number our napier constant). Therefore $$ \ln (x) = \log_e(x)$$ When you put $x=e$, we have $\ln(e)$, but that is simply $1$. Therefore $\big(\ln(e)\big)^2=1$.
$\endgroup$
