How do you simplify ln e^(2x)lne2x?

3 Answers
Aug 25, 2016

ln e^(2x) = 2xlne2x=2x

Explanation:

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

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

This is the definition of the Real natural logarithm.

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

e^(ln(e^t)) = e^teln(et)=et

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

ln e^t = tlnet=t

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

So if xx is any Real value:

ln e^(2x) = 2xlne2x=2x

May 31, 2017

2x2x

Explanation:

Using the property of logs:

log(a^b) = b log alog(ab)=bloga

We can see that:

ln(e^(2x))=2x ln eln(e2x)=2xlne

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

2xlne=2x2xlne=2x

May 31, 2018

2x2x

Explanation:

The key realization here is that lnxlnx and e^xex 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!