Does LNx cancle e^x? What is (lnx)^2?

Answer 1

well...ln(x) does not cancel e^x....but...

given e^x, if we take the natural log of e^x then we have ln(e^x), well this means that we have an exponent...we recall that we can pull the exponent down infront of a logrithm...so we now have xln(e)

ah! the ln(e) is 1, we recall that ln's are base e and if we take the logrithm of the base we get 1. so we have x*1=x

so...likewise...lnx^2 is 2ln(x)...and thats all we can do

hope this helps

matttlocke

Answer 2

ln(x) is another way to write "log base e of x"

If I had some number called (?), and I said that

(?) = ln(x)

I'm basically saying e to the (?) power is x.

If I had
(?) = ln(e^x), I'm saying e to the (?) power is e^x. Well duh, the (?) in this case could only be x. So yes, ln(e^x) is simply x.

Important: ln(1) is zero, and ln(e) is 1. Try to figure out why using the above examples.

Also, there is no special rule involving the square of a natural log.

Hope that helped,
UMRmathmajor

(p.s. These are important! You'll see them more in higher math.)

Answer 3

Yes they are mutually anti-functions, except that with ln(x), the natural log, x MUST be > 0.

Answer 4

I don't quite understand the question. ln(x) * e^x is not x, but ln(e^x) is.