# How do I solve #ln(1/2)^m = ln2#?

##### 1 Answer

Apr 1, 2017

Two ways...

One is simply recognizing that by algebra,

Another way is that anytime you see an exponent, try taking the

#ln(1/2)^(m) = ln2#

Use the property that

#lna^b = blna#

Thus, we get:

#mln(1/2) = ln2#

#mln(2)^(-1) = ln2#

#-mln2 = ln2#

#=> m = -1#