Question

for the log quetsion.
ln 1= 0 <-in this case, do i need to assume e is there?
i can't remember the formulas, is there any way i can remember it?

Answer 1

ln 1 means to what power must you raise e to get 1.

Answer 2

does that make sense?

Answer 3

yea, I guess i am misunderstanding form of questions. can you help me out with more please?

Answer 4

what are you having difficulty with?

Answer 5

ln e = 1. i am confuse with the formulas

Answer 6

what is the 1 stands for? is power?

Answer 7

yes, in that case it is the power of e.

Answer 8

i.e. ln e ^10 = 10.

Answer 9

ln and e are inverses of each other.

Answer 10

does that make sense?

Answer 11

yea, so that means above that example i show, ln e^1, is this right?

Answer 12

for ln e = 1 is just saying that the log of the base itself, which is e, is 1.

Answer 13

Exponential to Logarithmic:
\[b^{x} = a\]

I use b because b is the Base. B is for Base.

\[\log_{b} (a) = x\]

Fill in the blanks to change it to logarithmic using the numbers from the exponential form.

Always remember, A Log Is an Exponent. A log with Base "e" is a special log which isnt written as \[\log_{e} a\] , but rather it is written as \[\ln a\]

Answer 14

so in this case, In a, means log a without the base?

Answer 15

\[\ln a \] means \[\log_{e} a\] , but it just is written as \[\ln a \]

Answer 16

so for example, if ln 9, what is the answer going to be?

Answer 17

Therefore when you have ln e = 1
you actually have:
\[\log_{e} e = 1\]

Which, if you convert it back to exponential form, you get:
\[e ^{1} = e\]

Answer 18

ln 9 = \[\log_{e} 9\] which can convert to \[e ^{x} = 9\]

Answer 19

for the exponent, is it possible to become a # instead of x ?

Answer 20

ln 1=0
1 = e^0