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.