Why is ln(e^-2) = -2.

Why doesn't it = e^-2??

ln (e^x) = x ln e = x * 1 = x

so what if it was ln5(e^-2). would the answer be -10?? thanks for your help:)

No, it is only because ln is base e that ln(e^x) = x

in general
loga (a^x) = x loga(a) = x*1 = x

log 5 of 5 is 1, but log5 (e) is not one

okay thanks

it is all because

b^logb (x) = x
so
5^log5 (5) = 5
so log5 (5) has to be 1

thanks damon!!!!!!

To understand why ln(e^-2) is equal to -2 and not e^-2, we need to recall some important properties of logarithms and exponential functions.

The natural logarithm function, denoted as ln(x), is the inverse of the exponential function e^x. This means that ln(e^x) = x for any real number x. In other words, when you take the natural logarithm of the exponential of a number, you get back the original number.

Now let's apply this property to the expression ln(e^-2). Since e is the base of the natural logarithm, ln(e^-2) can be rewritten as ln(e^(-2)). Applying the inverse property, we get -2 as the result:

ln(e^-2) = -2

On the other hand, the expression e^-2 represents the exponential of -2, which is equivalent to 1/(e^2). It is important to note that e^-2 and 1/(e^2) are not the same. The former is an exponentiation, while the latter is the reciprocal of an exponential function.

So, to summarize:

ln(e^-2) = -2 (applying the inverse property of logarithms)
e^-2 = 1/(e^2) (representing the reciprocal of an exponential function)

Therefore, ln(e^-2) is equal to -2, not e^-2.