How to solve $(\ln e)^2$

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

• Note the following things: $\ln(x^2)=2\ln(x)$, $\ln(e)=1$ and $1^2=1$ – JMoravitz Dec 6 '17 at 3:56

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:

$$n^x = m$$

$$\ln e^2=2$$ 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:

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

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

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

• 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. – JMoravitz Dec 6 '17 at 5:07
• @JMoravitz Done both things! – Ravi Prakash Dec 6 '17 at 5:58

$\ln e$ is 1. So $(\ln e ) ^ 2$ is one.

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