How do I solve logarithms?

ln lxl =1

2 Answers
Apr 20, 2018

Please see below.

Explanation:

Here,

#ln |x|=1=>log_e|x|=1#

We know that,

#log_ex=y<=>x=e^y,where, x > 0#

#:.log_e|x|=1=>|x|=e^1#

#i.e. |x|=e=>x=+-e#

Note:

#"ln (x) is undefined when"# # x<=0"#

So, #ln|x|=ln|+-e|=ln e=1#

Apr 20, 2018

#x=+-e#

Explanation:

We use the definition:

# log_b x = y iff b^y = x#

So for the given expression:

# ln|x|=1 iff e^1 = |x| #

# :. |x| = e => x =+-e#