natural log?

Question

i want to take the natural log of (2e^a)
ie: ln|2e^a|

what happens to the 2??
thanks.

you guys rock...it's like putting fresh meat in a pirana tank....

Answer 1

ln2e^a
=ln2 + lne^a............since lnxy = lnx + lny
=ln2 + a.................since lne^x = x

the end
.

Answer 2

Hey there!

Here's the answer.

ln|2e^a| --> Write the problem.
ln|2|+ln|e^a| --> Use the product property of logarithms i.e. ln(mn)=ln m+ln n.
ln|2|+aln|e| --> Use the power property of logarithms i.e. ln(m^n)=n*ln(m).
ln|2|+a(1) --> Recall that ln(e)=1.
ln|2|+a --> Simplify the above expression.

So the answer is a+ln(2). The 2 is not a normal 2, but instead it is being natural "logarithmized" i.e. think of e^x=2, which is the same thing as x=ln(2).

Hope it helps!

Answer 3

ln (2e^a)
= ln 2 + ln e^a
= ln 2 + a ln e
= ln 2 + a
= 0.693 + a

Answer 4

I believe the 2 can be brought out front, if i remember the "ship jumper property" as i was taught were u just bring it out in front like multiplication.

Answer 5

ln(2.e^a)
=ln2 + lne^a
=ln2 + a lne
=ln2 + a

Answer 6

ln 2=0.7182.....