# Natural Log and e confusion

1. Aug 6, 2008

### Iron_Brute

I am starting my first year of college and I reviewing my high school notes from trig pre-cal and there was one thing I couldnt figure out. It was a multiple choice question and I dont have the text book anymore but the answer I circled I cant understand how I arrived to that answer.

1. The problem statement, all variables and given/known data
Given that x> 0 and y> 0, simplify e^ln(x) + ln(y)

3. The attempt at a solution
What I did was:
e^ln(x+y)
e^x+y
and I circled the answer: X+Y, but I get e^x+y which as an answer. All I wrote was ln and e cancel but I dont understand how they cancel.

2. Aug 6, 2008

### snipez90

Do you mean e^(ln(x) + ln(y)) ? If that's the case, then you add the two logarithms to get ln(xy) since the arguments are multiplied in log addition. Then using the fact that e^ln(x) = x, you have e^(ln(xy)) = xy. I'm not sure if that answers your question.

To see why e^ln(x) = x. Remember that ln(x) = log base e of x. Let ln(x) = y. Then converting to exponentiation gives e^y = x, but y = ln(x). Hence e^ln(x) = x.

3. Aug 6, 2008

### HallsofIvy

Do you mean e^(ln(x)+ ln(y))? What you wrote is e^(ln(x))+ ln(y).

No, ln(x)+ ln(y) is not equal to ln(x+ y). As snipez90 said, ln(x)+ ln(y)= ln(xy).
Then, of course, e^(ln(xy))= xy since e^x and ln(x) are inverse functions.

Another way to do that is to use the fact that e^(a+b)= e^a e^b:
e^(ln(x)+ ln(y))= (e^ln(x))(e^ln(y))= (x)(y)= xy.

4. Aug 7, 2008

### Iron_Brute

Thanks for the help. I see where I made my mistake now. I was using the wrong log identities, and misunderstand certain things about natural logs.