How do you simplify #ln e^(2x)#?

3 Answers
Aug 25, 2016

Answer:

#ln e^(2x) = 2x#

Explanation:

As a Real valued function, #x |-> e^x# is one to one from #(-oo, oo)# onto #(0, oo)#.

As a result, for any #y in (0, oo)# there is a unique Real value #ln y# such that #e^(ln y) = y#.

This is the definition of the Real natural logarithm.

If #t in (-oo, oo)# then #y = e^t in (0, oo)# and from the above definition:

#e^(ln(e^t)) = e^t#

Since #x |-> e^x# is one to one, we can deduce that for any Real value of #t#:

#ln e^t = t#

In other words, #t |-> e^t# and #t |-> ln t# are mutual inverses as Real valued functions.

So if #x# is any Real value:

#ln e^(2x) = 2x#

May 31, 2017

Answer:

#2x#

Explanation:

Using the property of logs:

#log(a^b) = b log a#

We can see that:

#ln(e^(2x))=2x ln e#

And since #ln(e) = log_e(e)=1#,

#2xlne=2x#

May 31, 2018

Answer:

#2x#

Explanation:

The key realization here is that #lnx# and #e^x# are inverses of each other, which cancel each other out. So we essentially have

#cancel(ln)cancel(e)^(2x)#

which just leaves us with #color(blue)(2x)#.

Hope this helps!